Chapter 3 — Functions and Graphs

Answer: C

Explanation

Step 1 — Substitute $x = -1$ carefully.

$$f(-1) = 2(-1)^2 + 3(-1) - 4$$

Step 2 — Evaluate each term.

$(-1)^2 = 1$, so $2(1) = 2$.

$3(-1) = -3$.

$$f(-1) = 2 + (-3) + (-4) = 2 - 3 - 4 = -5$$

Step 3 — Check each distractor's origin.

B) $-9$: The student treats $(-1)^2 = -1$ instead of $+1$, giving $2(-1) - 3 - 4 = -9$. Squaring always produces a non-negative result — $(-1)^2 = (-1)(-1) = +1$.

A) $1$: The student substitutes $x = 1$ (ignoring the negative), giving $2 + 3 - 4 = 1$.

D) $3$: The student makes a sign error on the constant, reading $-4$ as $+4$: $2 - 3 + 4 = 3$.

Takeaway: When the input is negative, write it in brackets — $(-1)^2$ — before simplifying. Never omit the brackets; $-1^2 = -(1^2) = -1$, but $(-1)^2 = +1$.

Answer: A

Explanation

Replace every $x$ with $(a+1)$, then expand:

$$f(a+1) = 5(a+1) + 3 = 5a + 5 + 3 = 5a + 8$$

Why the others are wrong:

• B ($5a+3$) evaluated $f(a)$ — the $+1$ was ignored.
• C ($5a+4$) wrote $5a + 1 + 3$, forgetting to multiply the 1 by 5.
• D ($5a+11$) treated the function as additive: $f(a) + f(1)$. Functions do not split across addition.

Takeaway: $f(\square) = 5(\square) + 3$ — the box holds the whole new expression, brackets and all.

Answer: D

Explanation

The rule: a relation is a function when every input has exactly one output.

• A — input 1 gives both 2 and 3. Not a function.
• B — $x^2+y^2=9$ is a circle; $x=0$ gives $y=\pm3$. Not a function.
• C — the vertical line $x=3$ sends one input to every $y$ at once. Not a function.
• D — $1\to5$, $2\to5$, $3\to7$. Each input has one output ✓ Two inputs sharing an output is fine.

Takeaway: Two inputs may share an output; one input may never have two. That is exactly what the vertical line test checks — if a vertical line crosses the graph twice, it is not a function.

Answer: B

Explanation

$g(f(x))$ means apply $f$ first, then $g$. So the input to $g$ is $2x+1$:

$$g(f(x)) = (2x+1)^2 = 4x^2 + 4x + 1$$

Why the others are wrong:

• A ($4x^2+1$) dropped the middle term. A binomial square always has three terms.
• D ($2x^2+1$) computed $f(g(x))$ instead — composition is not commutative.
• C ($4x^2+2x+1$) used $a^2+ab+b^2$, missing the factor of 2.

Takeaway: $(a+b)^2 = a^2 + 2ab + b^2$ — the middle term is $2ab$. Read $g(f(x))$ as "put $f(x)$ into $g$", working inside out.

Answer: A

Explanation

Substitute $-x$ and compare with $f(x)$:

$$f(-x) = (-x)^4 - 3(-x)^2 = x^4 - 3x^2 = f(x)$$

Since $f(-x) = f(x)$, the function is even.

Why the others are wrong:

• B (odd) would need $f(-x) = -f(x)$, i.e. $-x^4+3x^2$. Not equal.
• C claims mixed exponents. Both exponents here (4 and 2) are even — which is exactly why it works.
• D says polynomials cannot be even or odd. They are the commonest examples: $f(x)=x^2$ is the simplest even one.

Takeaway: Even exponents throughout → even function. One odd-exponent term among them makes the function neither.

Answer: D

Explanation

Step 1 — Set up the equation.

$$3x - 7 = 14$$

Step 2 — Solve.

$$3x = 21 \implies x = 7$$

Step 3 — Explain each error.

A) $\dfrac{14}{3}$: The student solves $3x = 14$ (ignores the $-7$, treating the equation as if $g(x) = 3x$).

C) $3$: The student correctly computes $3x = 21$ but then divides $21$ by $7$ (the answer) instead of by $3$: $21 \div 7 = 3$.

B) $63$: The student correctly gets $3x = 21$ but then multiplies rather than divides: $21 \times 3 = 63$.

Takeaway: Always isolate the variable by doing the same operation to both sides. After adding 7: $3x = 21$. Dividing both sides by 3 gives $x = 7$, not $x = 21 \times 3$.

Answer: C

Explanation

Choose the right piece for each input. $-2 < 0$ uses the first; $3 \geq 0$ uses the second.

$$f(-2) = (-2)^2 - (-2) = 4 + 2 = 6 \qquad f(3) = 2(3)+1 = 7$$ $$f(-2) + f(3) = 13$$

Why the others are wrong:

• A (3) used the wrong piece for both inputs.
• B (9) slipped a sign: read $-(-2)$ as $-2$, giving $f(-2)=2$.
• D (12) dropped the $+1$ from $f(3)$.

Takeaway: Write down which piece you are using before you substitute. One extra second prevents the commonest error here.

Section 2 — Domain and Range

The domain is the set of all valid inputs; the range is the set of all possible outputs. Two rules govern domain restrictions for standard functions: you cannot take the square root of a negative number (so the expression under a root must be $\geq 0$), and you cannot divide by zero (so the denominator must be $\neq 0$). Logarithms add a third: the argument must be strictly positive. For the range, ask what values the output can possibly reach — often easiest by thinking about the graph.

Answer: B

Explanation

Step 1 — Require the expression under the root to be non-negative.

$$x - 3 \geq 0 \implies x \geq 3$$

Step 2 — Check the boundary.

At $x = 3$: $\sqrt{3 - 3} = \sqrt{0} = 0$. This is defined, so $x = 3$ is included. The domain uses $\geq$, not $>$.

Step 3 — Explain errors.

A) $x > 3$: Excludes $x = 3$, but $\sqrt{0} = 0$ is perfectly defined. The strict inequality is wrong.

D) $x \leq 3$: Reverses the inequality entirely.

C) All real numbers: Ignores the restriction. $f(2) = \sqrt{-1}$, which is not real.

Takeaway: Square-root domains use $\geq 0$ (not $> 0$), because zero is a valid input: $\sqrt{0} = 0$. Only the square root of a strictly negative number is undefined in the real numbers.

Answer: D

Explanation

Step 1 — Identify when the denominator equals zero.

$$x - 4 = 0 \implies x = 4$$

Step 2 — State the domain.

All real numbers are valid inputs except $x = 4$, which makes the denominator zero (division by zero is undefined).

Step 3 — Trace errors.

B) $x > 4$: Excludes all values less than 4 unnecessarily — $f(0) = \frac{1}{-4} = -\frac{1}{4}$ is perfectly defined.

C) $x \geq 4$: Excludes values $< 4$ and still includes the problematic $x = 4$.

A) All real numbers: Does not account for $x = 4$, where the function is undefined.

Takeaway: A rational function is undefined only where the denominator equals zero. Every other real number is a valid input, including negatives and numbers less than 4.

Answer: C

Explanation

Step 1 — Find the minimum of $x^2 + 2$.

$x^2 \geq 0$ for all real $x$, with minimum value $0$ at $x = 0$.

Adding 2: $f(x) = x^2 + 2 \geq 0 + 2 = 2$.

Step 2 — Check whether the minimum is achieved.

At $x = 0$: $f(0) = 0 + 2 = 2$. The value 2 is achieved, so the range includes 2.

$$\text{Range: } y \geq 2$$

Step 3 — Explain errors.

A) $y \geq 0$: Ignores the $+2$ shift — this would be the range of $y = x^2$, not $y = x^2 + 2$.

B) All real numbers: Would require negative outputs, but $x^2 + 2 \geq 2 > 0$ always.

D) $y > 2$: Excludes $y = 2$, but $f(0) = 2$, so 2 is in the range.

Takeaway: The range of $f(x) = x^2 + c$ is $y \geq c$, because $x^2 \geq 0$ and the shift $c$ moves the minimum up to $c$.

Answer: A

Explanation

Step 1 — Require the radicand to be non-negative.

$$4 - x^2 \geq 0$$

Step 2 — Solve the inequality.

$$x^2 \leq 4 \implies |x| \leq 2 \implies -2 \leq x \leq 2$$

Step 3 — Verify the boundary.

At $x = \pm 2$: $4 - 4 = 0 \geq 0$. Both endpoints are valid.

Step 4 — Explain errors.

C) $x \geq -2$: Captures the left boundary correctly but misses the right: for $x = 3$, $4 - 9 = -5 < 0$ — undefined.

B) $x \leq 2$: Captures the right boundary but misses the left: for $x = -3$, $4 - 9 = -5 < 0$ — undefined.

D) $x \geq 0$: Restricts to non-negative inputs only, losing the left half of the valid domain.

Takeaway: When the radicand is a downward-opening expression $c - x^2$, the domain is a closed interval $[-\sqrt{c},\, \sqrt{c}]$. Always solve both sides of the inequality simultaneously.

Answer: B

Explanation

The form is $a(x-p)^2+q$ with $a=-1$, vertex $(1, 9)$.

Since $a < 0$ the parabola opens downward, so 9 is a maximum:

$$\text{Range: } y \leq 9$$

Why the others are wrong:

• A ($y \geq 9$) treats the vertex as a minimum. Downward-opening means maximum.
• D ($y \geq 0$) describes an upward parabola with vertex at the origin — two errors at once.
• C (all reals) would need arbitrarily large values, but the expression never exceeds 9.

Takeaway: The sign of $a$ decides the range: $a>0$ gives $y \geq q$, $a<0$ gives $y \leq q$.

Answer: D

Explanation

A logarithm needs a strictly positive argument:

$$2x - 6 > 0 \implies 2x > 6 \implies x > 3$$

Why the others are wrong:

• B ($x>6$) reached $2x>6$ then forgot to divide by 2.
• C ($x \geq 3$) includes $x=3$, where the argument is 0 — and $\ln(0)$ is undefined.
• A ($x>0$) restricts $x$ instead of the argument. The condition comes from $2x-6$, not $x$.

Takeaway: For logs, the argument must be $> 0$ — strictly. Never $\geq 0$.

Answer: A

Explanation

Domain: the denominator $(x-2)^2$ is zero only at $x=2$, so $x \neq 2$.

Range: a square is positive everywhere except at $x=2$ (excluded), so $f(x) = \frac{1}{(x-2)^2}$ is always positive. It can never equal 0 — that would need $1=0$. And it reaches every positive value: it grows without bound near $x=2$ and shrinks towards 0 far away. So $y > 0$.

Why the others are wrong:

• B (domain $x>2$) excludes the left side, but $f(0) = \frac14$ is perfectly fine.
• C (range $y \geq 0$) includes 0, which $f$ never reaches.
• D (all reals) ignores $x=2$.

Takeaway: For $\frac{1}{(x-p)^2}$: domain $x \neq p$, range $y > 0$. A squared denominator makes the output strictly positive and never zero.

Section 3 — Straight-Line Functions

A straight line in the form $y = mx + c$ is fully described by its gradient $m$ and its $y$-intercept $c$. The gradient measures steepness and direction: $m = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}$. Two lines are parallel if and only if they have equal gradients; perpendicular if the product of their gradients equals $-1$.

Answer: C

Explanation

Step 1 — Apply the gradient formula.

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{0 - (-2)} = \frac{4}{2} = 2$$

Step 2 — Explain errors.

A) $4$: Student reads off the $y$-intercept (where the line crosses the $y$-axis at $(0,4)$) and reports it as the gradient.

B) $-2$: Student reads off the $x$-intercept (the line crosses the $x$-axis at $(-2,0)$) and reports it as the gradient.

D) $\dfrac{1}{2}$: Student inverts the fraction: $\dfrac{x_2 - x_1}{y_2 - y_1} = \dfrac{2}{4} = \frac{1}{2}$ (run over rise, not rise over run).

Takeaway: Gradient = rise $\div$ run = $\dfrac{\Delta y}{\Delta x}$. Choose your two points consistently: $y_2 - y_1$ on top, $x_2 - x_1$ on the bottom, using the same subscript order.

Answer: B

Explanation

Step 1 — Use point-gradient form.

$$y - y_1 = m(x - x_1)$$ $$y - 4 = 3(x - 2)$$

Step 2 — Expand.

$$y - 4 = 3x - 6$$ $$y = 3x - 6 + 4 = 3x - 2$$

Step 3 — Explain errors.

A) $y = 3x + 4$: Student uses the $y$-coordinate of the given point as the $y$-intercept directly: $c = 4$. This is only valid when the given point is the $y$-intercept (i.e., when $x_1 = 0$).

D) $y = 3x - 4$: Student computes $c = y_1 - x_1 = 4 - 2 \times \ldots$ or makes an arithmetic error in the expansion.

C) $y = 3x + 2$: Student uses $c = x_1 = 2$ — reads the $x$-coordinate as the intercept.

Takeaway: The $y$-intercept $c$ is found by substituting the given point into $y = mx + c$ and solving: $4 = 3(2) + c \Rightarrow c = 4 - 6 = -2$.

Answer: A

Explanation

Gradient: $m = \dfrac{9-3}{2-(-1)} = \dfrac{6}{3} = 2$

Intercept: substitute $(-1, 3)$ into $y = 2x + c$: $$3 = 2(-1) + c \implies c = 5$$ $$y = 2x + 5$$

Check with the other point: $2(2)+5 = 9$ ✓

Why the others are wrong:

• B ($c=7$) subtracted the gradient from $y$ instead of $2x_2 = 4$.
• C ($c=3$) read $c$ straight off the first point's $y$-coordinate.
• D ($c=1$) used $x_1 = +1$ instead of $-1$, losing the sign.

Takeaway: Substitute both coordinates into $y = mx+c$ and watch the signs — here $2(-1) = -2$, not $+2$.

Answer: D

Explanation

Step 1 — Recall the condition for parallel lines.

Two lines are parallel if and only if they have the same gradient.

The given line $y = 3x - 7$ has gradient $m = 3$.

Step 2 — Find each option's gradient.

A) $y = 3x + 7$: gradient $= 3$. Parallel.

D) $3y = x + 5 \implies y = \dfrac{1}{3}x + \dfrac{5}{3}$: gradient $= \dfrac{1}{3} \neq 3$. NOT parallel. ✓

C) $y - 3x = 2 \implies y = 3x + 2$: gradient $= 3$. Parallel.

B) $2y = 6x - 1 \implies y = 3x - \dfrac{1}{2}$: gradient $= 3$. Parallel.

Takeaway: Always rearrange to $y = mx + c$ before reading the gradient. A coefficient in front of $y$ (like $3y$ or $2y$) changes the gradient by a factor — divide both sides first.

Answer: B

Explanation

Step 1 — Apply the perpendicularity condition.

If two lines are perpendicular, the product of their gradients is $-1$:

$$m_1 \times m_2 = -1$$

Step 2 — Solve for $m_2$.

$$\frac{1}{4} \times m_2 = -1 \implies m_2 = -4$$

Step 3 — Explain errors.

A) $\dfrac{1}{4}$: Same gradient — this describes a parallel line, not a perpendicular one.

C) $4$: The reciprocal without the sign change. The perpendicular gradient is the negative reciprocal.

D) $-\dfrac{1}{4}$: The negative without the reciprocal. Two separate operations are needed: flip the fraction AND change the sign.

Takeaway: Perpendicular gradient = $-\dfrac{1}{m}$. For $m = \frac{1}{4}$: flip to get $\frac{4}{1} = 4$, then negate to get $-4$.

Answer: C

Explanation

Step 1 — Set $y = 0$ (the $x$-intercept lies on the $x$-axis, where $y = 0$).

$$3x - 4(0) = 24$$ $$3x = 24 \implies x = 8$$

Step 2 — Explain errors.

A) $-6$: Student finds the $y$-intercept by setting $x = 0$: $-4y = 24 \Rightarrow y = -6$ — this is the $y$-intercept, not the $x$-intercept.

B) $6$: Student finds the $y$-intercept ($y = -6$) but takes its absolute value.

D) $-8$: Student applies a sign error when dividing: $3x = 24 \Rightarrow x = -8$ (incorrectly negates).

Takeaway: For intercepts, remember the rule: $x$-intercept → set $y = 0$; $y$-intercept → set $x = 0$. They are different points and will usually have different values.

Answer: B

Explanation

Set the two expressions equal and solve: $$3x - 1 = x + 5 \implies 2x = 6 \implies x = 3$$

Then find $y$: $y = 3(3) - 1 = 8$, so the point is $(3, 8)$.

Check in the other equation: $3 + 5 = 8$ ✓

Why the others are wrong:

• A $(3, 14)$ found $x=3$ then mixed the two equations when computing $y$.
• C $(6, 17)$ reached $2x = 6$ but read off $x = 6$, forgetting to divide.
• D $(2, 7)$ made a sign slip: $5 - 1$ instead of $5 + 1$.

Takeaway: Equate the two $y$-expressions, solve for $x$, substitute back — then check in the other equation.

Section 4 — Parabolas

The parabola $y = a(x - p)^2 + q$ (vertex form) has its vertex at $(p, q)$ and opens upward when $a > 0$, downward when $a < 0$. From standard form $y = ax^2 + bx + c$: the axis of symmetry is $x = -\dfrac{b}{2a}$, the discriminant $\Delta = b^2 - 4ac$ tells you the number of $x$-intercepts, and the $y$-intercept is simply $c$.

Answer: A

Explanation

Step 1 — Find the $x$-coordinate of the vertex.

$$x = -\frac{b}{2a} = -\frac{-6}{2(1)} = \frac{6}{2} = 3$$

Step 2 — Find the $y$-coordinate.

$$y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$$

Vertex: $(3, -4)$.

Step 3 — Explain errors.

B) $(6, 5)$: Student reads off the values of $b$ and $c$ directly as the vertex coordinates — these are not vertex coordinates.

C) $(3, 4)$: Student correctly finds $x = 3$ but takes the absolute value of $y$: $|-4| = 4$, ignoring the negative sign.

D) $(-3, 4)$: Student negates the $x$-coordinate (perhaps thinking $x = -b/2a = -3$) and takes $|y| = 4$.

Takeaway: The $y$-coordinate of the vertex must be calculated by substituting $x = 3$ back into the original equation. Do not read it from the coefficients.

Answer: D

Explanation

Factorise: two numbers multiplying to $+6$ and adding to $-5$ are $-2$ and $-3$. $$x^2 - 5x + 6 = (x-2)(x-3) = 0 \implies x = 2 \text{ or } 3$$

Why the others are wrong:

• A ($1, 6$) multiply to 6 ✓ but add to 7 ✗ — factor pairs guessed without checking the sum.
• B ($-2, -3$) come from $(x+2)(x+3) = x^2+5x+6$ — wrong signs throughout.
• C ($-1, 5$) add to 4, not $-5$.

Takeaway: Both conditions must hold: product $= c$ and sum $= b$. Product $+6$ with sum $-5$ forces both numbers negative.

Answer: C

Explanation

Step 1 — Read the vertex form $y = a(x - p)^2 + q$.

Rewrite as $y = -1 \cdot (x - (-2))^2 + 8$.

So $a = -1$, $p = -2$, $q = 8$. Vertex: $(-2, 8)$.

Step 2 — Determine direction.

$a = -1 < 0$: the parabola opens downward.

Step 3 — Explain errors.

A) Vertex $(2, 8)$: Student reads $+2$ as the $p$-value, ignoring that $y = a(x - p)^2 + q$ requires $x - p$; since we have $x + 2 = x - (-2)$, $p = -2$.

B) Vertex $(-2, -8)$: Vertex $x$-coordinate is correct, but student negates the $q$ value: $-8$ instead of $8$.

D) Vertex $(-2, 8)$: Both vertex coordinates are correct but direction is wrong — a negative $a$ always gives a downward opening parabola.

Takeaway: In $y = a(x - p)^2 + q$, the vertex is $(p, q)$. Watch the sign: $(x + 2) = (x - (-2))$, so $p = -2$, not $+2$.

Answer: B

Explanation

Step 1 — Apply the formula.

$$x = -\frac{b}{2a} = -\frac{-8}{2(2)} = \frac{8}{4} = 2$$

Step 2 — Explain errors.

A) $x = -2$: Student applies the formula as $\dfrac{b}{2a} = \dfrac{-8}{4} = -2$, forgetting the leading negative sign in $-\dfrac{b}{2a}$. Since $b = -8$, we need $-(-8) = +8$ in the numerator.

C) $x = 4$: Student uses $\dfrac{b}{a} = \dfrac{-8}{2} = -4$... or perhaps $\dfrac{|b|}{a} = \dfrac{8}{2} = 4$ (omits the denominator factor of 2).

D) $x = -4$: Student uses $\dfrac{b}{a} = \dfrac{-8}{2} = -4$ (omits the factor of 2 in $2a$, and keeps the sign).

Takeaway: Memorise $x = -\dfrac{b}{2a}$ as a formula — the numerator is $-b$, the denominator is $2a$. For $2x^2 - 8x + 3$: $a = 2$, $b = -8$, so $x = -(-8) / (2 \times 2) = 8/4 = 2$.

Answer: C

Explanation

Start from vertex form with vertex $(2,-3)$: $\ y = a(x-2)^2 - 3$

Substitute the point $(4,5)$ to find $a$: $$5 = a(2)^2 - 3 = 4a - 3 \implies a = 2$$ $$y = 2(x-2)^2 - 3$$

Why the others are wrong:

• A assumed $a=1$ without checking. At $x=4$ it gives $y=1$, not 5.
• B uses $(x+2)$, putting the vertex at $-2$.
• D uses $+3$, putting the vertex at $y=3$.

Takeaway: The vertex gives you $p$ and $q$ for free, but $a$ needs one extra point. Always substitute it and solve.

Answer: A

Explanation

Step 1 — Identify vertex and direction.

Vertex: $(-3, 4)$. Leading coefficient $a = -1 < 0$: opens downward.

Step 2 — State the range.

The vertex gives the maximum value of 4. The parabola extends downward without bound:

$$\text{Range: } y \leq 4$$

Step 3 — Explain errors.

C) $y \geq 4$: Treats the vertex as a minimum — correct logic for upward parabolas, wrong here.

B) $y \leq -4$: Negates the $q$ value incorrectly. The vertex height is $+4$, not $-4$.

D) $y \geq -4$: Combines both errors: negates $q$ and treats it as a minimum.

Takeaway: Range of $-(x-p)^2 + q$: the maximum is $q$ (at the vertex), so range = $y \leq q$.

Answer: D

Explanation

Step 1 — Set $x = 0$.

$$y = 3(0)^2 - 2(0) + 7 = 0 - 0 + 7 = 7$$

Step 2 — Explain errors.

A) $y = 0$: Student sets $y = 0$ (finding the $x$-intercept) rather than $x = 0$.

B) $y = 5$: Student evaluates $-2x$ at $x = 0$ as $-2$ (forgetting to multiply by $0$): $0 - 2 + 7 = 5$. The term $-2x$ at $x = 0$ is $-2(0) = 0$, not $-2$.

C) $y = -7$: Student negates the constant term, reading $+7$ as $-7$.

Takeaway: The $y$-intercept of any polynomial $y = ax^n + \ldots + c$ is always the constant term $c$, found by setting $x = 0$. Every term with $x$ vanishes.

Answer: A

Explanation

Step 1 — Compute the discriminant.

$$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(7) = 16 - 28 = -12$$

Step 2 — Interpret.

$\Delta < 0$: no real roots. The parabola does not intersect the $x$-axis.

Step 3 — Explain errors.

B) $1$ real root: Would require $\Delta = 0$ (the parabola touches the $x$-axis exactly once).

C) $2$ distinct real roots: Would require $\Delta > 0$.

D) Cannot be determined: The discriminant gives a definitive answer — no graph required.

Takeaway: $\Delta < 0 \Rightarrow$ no real roots; $\Delta = 0 \Rightarrow$ one repeated root; $\Delta > 0 \Rightarrow$ two distinct real roots.

Answer: B

Explanation

Step 1 — Factor.

Find two numbers with product $-8$ and sum $-2$: these are $-4$ and $+2$.

$$x^2 - 2x - 8 = (x - 4)(x + 2)$$

Step 2 — Solve.

$$x = 4 \quad \text{or} \quad x = -2$$

Step 3 — Explain errors.

A) $x = 4$ and $x = 2$: Uses $(x-4)(x-2) = x^2 - 6x + 8$ — wrong constant term.

C) $x = -4$ and $x = 2$: Uses $(x+4)(x-2) = x^2 + 2x - 8$ — correct constant but wrong sign on the middle term (sum is $+2$, not $-2$).

D) $x = -4$ and $x = -2$: Uses $(x+4)(x+2) = x^2 + 6x + 8$ — wrong on both middle and constant terms.

Takeaway: Always check both the product and the sum of your chosen factor pair: product $= c$, sum $= b$. A quick multiplication of the factors confirms the factorisation before you state the roots.

Section 5 — Hyperbolas

The standard hyperbola is $y = \dfrac{a}{x}$ or its translated form $y = \dfrac{a}{x + p} + q$. The vertical asymptote is at $x = -p$ (where the denominator equals zero) and the horizontal asymptote is at $y = q$ (the additive shift). When $a > 0$, the graph occupies Quadrants 1 and 3; when $a < 0$, it occupies Quadrants 2 and 4.

Answer: D

Explanation

Step 1 — Find the vertical asymptote.

The denominator is $x$. Setting $x = 0$ gives a zero denominator: vertical asymptote at $x = 0$ (the $y$-axis).

Step 2 — Find the horizontal asymptote.

As $x \to \pm\infty$, $y = \frac{2}{x} \to 0$: horizontal asymptote at $y = 0$ (the $x$-axis).

Step 3 — Explain errors.

A) $x = 2$ and $y = 0$: Student reads the numerator $a = 2$ as the vertical asymptote — but asymptotes come from the denominator (VA) and the shift (HA), not the numerator.

B) $x = 0$ and $y = 2$: Correct VA, but student reads the numerator as the HA.

C) No asymptotes: Every rational function of this form has exactly two asymptotes.

Takeaway: For $y = \dfrac{a}{x}$: VA = $x = 0$, HA = $y = 0$. The numerator $a$ affects the stretch and which quadrants are occupied — it does not move the asymptotes.

Answer: C

Explanation

Step 1 — Set the denominator equal to zero.

$$x - 2 = 0 \implies x = 2$$

Step 2 — Explain errors.

A) $x = 1$: Student reads the horizontal shift $q = 1$ as the vertical asymptote.

B) $x = -2$: Student negates the value from the denominator, perhaps thinking of $x + 2 = 0$ instead of $x - 2 = 0$.

D) $x = 3$: Student reads the numerator $a = 3$ as the asymptote.

Takeaway: Vertical asymptote of $y = \dfrac{a}{x - p} + q$: set $x - p = 0$, giving $x = p$. The other numbers ($a$ and $q$) do not determine the VA.

Answer: D

Explanation

Step 1 — Apply the rule.

For $y = \dfrac{a}{x + p} + q$, as $x \to \pm\infty$, the fraction $\dfrac{a}{x+p} \to 0$, so $y \to q$.

Here $q = -4$: horizontal asymptote $y = -4$.

Step 2 — Explain errors.

B) $y = 4$: Student drops the sign from $q = -4$.

C) $y = -3$: Student reads the vertical-asymptote value $p = -3$ (giving $x = -3$) as a $y$-value.

A) $y = 0$: Applies the un-shifted rule $y = \frac{a}{x}$, where HA is $y = 0$. This ignores the $-4$ shift.

Takeaway: For $y = \dfrac{a}{x + p} + q$: VA at $x = -p$, HA at $y = q$. The $+q$ shifts the entire graph — including the horizontal asymptote.

Answer: A

Explanation

Step 1 — Find where the denominator equals zero.

$$x + 5 = 0 \implies x = -5$$

Domain: all real $x$ except $x = -5$.

Step 2 — Explain errors.

B) $x \neq 5$: Student reads $x + 5 = 0$ as $x = 5$ (drops the sign change on rearrangement).

C) $x > -5$: Only excludes values to the left of $-5$ unnecessarily — $f(-6) = \frac{1}{-1} = -1$ is defined.

D) All real numbers: Ignores the restriction from the zero denominator.

Takeaway: $x + 5 = 0 \Rightarrow x = -5$. The domain is all reals except the asymptote $x = -5$, written $x \neq -5$ or $\mathbb{R} \setminus \{-5\}$.

Answer: C

Explanation

Step 1 — Test the sign of $y$ in each quadrant.

When $x > 0$: $y = \dfrac{2}{x} > 0$ (positive ÷ positive = positive) → Quadrant 1.

When $x < 0$: $y = \dfrac{2}{x} < 0$ (positive ÷ negative = negative) → Quadrant 3.

Step 2 — Explain errors.

A) Quadrants 1 and 2: In Quadrant 2, $x < 0$, so $y < 0$ — not Quadrant 2.

B) Quadrants 2 and 4: This is the pattern for $y = \dfrac{-2}{x}$ (negative numerator).

D) All four quadrants: The graph never crosses the axes (asymptotes at $x = 0$ and $y = 0$), so it cannot be in all four.

Takeaway: Sign of $a$ in $y = \dfrac{a}{x}$ determines the quadrants: $a > 0$ → Quadrants 1 and 3 (where $x$ and $y$ have the same sign); $a < 0$ → Quadrants 2 and 4.

Answer: B

Explanation

The whole output has been negated: $y = -\left(\dfrac{3}{x}\right)$. Negating the output is the definition of a reflection in the $x$-axis — every point $(x,y)$ becomes $(x,-y)$.

On the graph, this moves the two branches out of Quadrants 1 and 3 and into Quadrants 2 and 4.

Why the others are wrong:

• A — a vertical shift would give $\frac{3}{x} - 6$, which has a horizontal asymptote at $y=-6$. Nothing here shifted.
• C — reflecting in $y=x$ gives the inverse. Swapping $x$ and $y$ in $y=\frac3x$ returns $y=\frac3x$ itself, so that reflection changes nothing.
• D — a vertical stretch would give $\frac{9}{x}$. The graph has been flipped, not stretched.

Takeaway: $y = -f(x)$ is always a reflection in the $x$-axis. A curiosity worth noticing: for this particular hyperbola, reflecting in the $y$-axis produces the same picture, because $\frac{3}{-x}$ is also $-\frac3x$.

Answer: B

Explanation

Equate the two expressions and clear the fraction ($x \neq 0$): $$\frac{6}{x} = x+1 \implies 6 = x^2 + x \implies x^2 + x - 6 = 0$$

Factorise: $(x+3)(x-2) = 0$, so $x = -3$ or $x = 2$.

Sum: $-3 + 2 = -1$

Why the others are wrong:

• A ($1$) has a sign error on $b$ in Vieta's sum, $-b/a$.
• C ($-6$) reports the product of the roots instead of the sum.
• D ($5$) added the absolute values, $3 + 2$.

Takeaway: Equate, clear fractions, solve. Vieta gives a shortcut: for $x^2+bx+c=0$ the roots sum to $-b$ and multiply to $c$ — so you can answer without factorising at all.

Section 6 — Exponential and Logarithmic Functions

Exponential functions have the form $y = a \cdot b^x$ (with $b > 0$, $b \neq 1$). The horizontal asymptote is always $y = 0$ unless a vertical shift is added. Logarithms are the inverses of exponentials: $y = \log_b x \iff b^y = x$. The key log laws — product, quotient, and power rules — convert between forms that look different but are equivalent.

Answer: D

Explanation

Step 1 — Compute.

$$f(5) = 2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$$

Step 2 — Explain errors.

A) $10$: Student multiplies instead of raising to a power: $2 \times 5 = 10$.

C) $25$: Student evaluates $5^2 = 25$ — swaps base and exponent.

B) $16$: Student evaluates $2^4 = 16$ — uses exponent 4 instead of 5.

Takeaway: $2^5$ means five factors of 2 multiplied together — not $2 \times 5$. Build the powers: $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32$.

Answer: A

Explanation

Step 1 — Consider the behaviour as $x \to -\infty$.

As $x \to -\infty$: $2^x \to 0^+$, so $3 \cdot 2^x \to 3 \cdot 0 = 0$.

The horizontal asymptote is $y = 0$.

Step 2 — Explain errors.

C) $y = 3$: The coefficient 3 is a vertical stretch — it scales the function but does not shift the asymptote. A vertical shift (i.e., $3 \cdot 2^x + k$) would shift the HA.

B) $y = 2$: The base 2 determines the growth rate, not the asymptote.

D) $y = 6$: Student multiplies the coefficient and the base: $3 \times 2 = 6$.

Takeaway: For $y = a \cdot b^x$ (with no added constant), the HA is always $y = 0$ — the coefficient $a$ only stretches the graph vertically, it does not move the asymptote.

Answer: C

Explanation

Step 1 — Express 64 as a power of 2.

$$2^1 = 2,\quad 2^2 = 4,\quad 2^3 = 8,\quad 2^4 = 16,\quad 2^5 = 32,\quad 2^6 = 64$$

$$2^x = 2^6 \implies x = 6$$

Step 2 — Explain errors.

A) $x = 32$: Student divides: $64 \div 2 = 32$, treating the problem as $2x = 64$.

B) $x = 8$: Student recalls $8^2 = 64$ and uses the exponent 2 or the base 8 — neither is correct here.

D) $x = 4$: Student confuses with $4^3 = 64$ or $\sqrt[3]{64} = 4$ — different base.

Takeaway: Rewrite both sides as powers of the same base. $64 = 2^6$, so $2^x = 2^6 \Rightarrow x = 6$. Memorise: $2^{10} = 1024$, $2^8 = 256$, $2^7 = 128$, $2^6 = 64$.

Answer: A

Explanation

Step 1 — Determine increasing or decreasing.

The base is $\dfrac{1}{2} < 1$. For exponentials $y = b^x$: if $b < 1$, the function is decreasing (as $x$ increases, you raise a fraction to a higher power, making it smaller).

Step 2 — Find the $y$-intercept.

At $x = 0$: $y = \left(\frac{1}{2}\right)^0 = 1$.

Step 3 — Explain errors.

B) Increasing: Would hold for $b > 1$ (e.g., $y = 2^x$). With base $\frac{1}{2} < 1$, the function decreases.

C) $y$-intercept $0$: Any expression $b^0 = 1$ regardless of $b$ (provided $b \neq 0$). The $y$-intercept of any basic exponential is 1.

D) $y$-intercept $2$: Student reads $\frac{1}{2}$ as the base and takes its reciprocal $2$ as the $y$-intercept — the $y$-intercept is not the reciprocal of the base.

Takeaway: $y = b^x$ has $y$-intercept $1$ for any base $b$. The base determines whether the function grows ($b > 1$) or decays ($0 < b < 1$).

Answer: B

Explanation

Step 1 — Convert to exponential form.

$$\log_2(x) = 4 \iff 2^4 = x$$

$$x = 16$$

Step 2 — Explain errors.

A) $x = 8 = 2^3$: Off by one in the exponent.

C) $x = 4$: Student uses $x = $ the RHS value (the logarithm result) rather than converting: $\log_2(4) = 2$, not 4.

D) $x = 2$: Student uses the base as the answer.

Takeaway: $\log_b(x) = n \iff x = b^n$. Here: $\log_2(x) = 4 \iff x = 2^4 = 16$.

Answer: D

Explanation

Step 1 — Apply the power rule.

$$\log_a(x^3) = 3 \log_a x$$

The power rule states: the exponent comes down as a multiplier in front of the logarithm.

Step 2 — Explain errors.

A) $(\log_a x)^3$: The student cubes the logarithm instead of multiplying by 3. $(\log_a x)^3 \neq 3 \log_a x$ unless $\log_a x = 3$ or $0$.

B) $\dfrac{3}{\log_a x}$: No logarithm law produces division by the logarithm.

C) $\log_a(3x)$: Student brings the exponent inside the argument as a coefficient: $x^3 \to 3x$. The product rule handles $\log_a(3x) = \log_a 3 + \log_a x$, which is unrelated.

Takeaway: Power rule: $\log_a(x^n) = n \log_a x$. The exponent moves to become a multiplier, not a factor inside the argument.

Answer: C

Explanation

Step 1 — Convert to exponential form.

$$\log_3(x + 1) = 2 \iff x + 1 = 3^2 = 9$$

Step 2 — Solve for $x$.

$$x = 9 - 1 = 8$$

Step 3 — Explain errors.

A) $x = 1$: Student writes $x + 1 = 2$ (converts $\log_3(\cdot) = 2$ as "the argument equals 2" instead of $3^2$).

B) $x = 9$: Student correctly reaches $x + 1 = 9$ but forgets to subtract 1, reporting $x = 9$.

D) $x = 6$: Student computes $3 \times 2 = 6$ (multiplies base by exponent instead of raising base to exponent).

Takeaway: Convert $\log_b(A) = n$ to $A = b^n$ first, then solve. Here: $x + 1 = 9 \Rightarrow x = 8$.

Answer: C

Explanation

Step 1 — Find the inverse by swapping $x$ and $y$.

$$x = 3^y$$

Step 2 — Solve for $y$ by converting to logarithm form.

$$y = \log_3 x$$

Step 3 — Explain errors.

B) $y = x^3$: Confuses $3^x$ (exponential) with $x^3$ (power function). Their inverses are $\log_3 x$ and $x^{1/3}$ respectively.

A) $y = 3^{-x}$: This is a reflection of $y = 3^x$ across the $y$-axis, not its inverse.

D) $y = \dfrac{x}{3}$: Applies a linear-function inverse strategy (divide by 3) to an exponential.

Takeaway: Exponentials and logarithms are inverses: $f(x) = b^x \iff f^{-1}(x) = \log_b x$. To find the inverse: swap $x$ and $y$, then use the definition of logarithm to isolate $y$.

Answer: B

Explanation

Step 1 — Set $x = 0$.

$$y = 2^{0+1} = 2^1 = 2$$

Step 2 — Explain errors.

A) $y = 1$: Student evaluates $2^0 = 1$ (ignores the $+1$ in the exponent, as if the function were $y = 2^x$).

C) $y = 3$: Student adds: $2 + 1 = 3$ instead of evaluating $2^1$.

D) $y = 0$: Student confuses with the $x$-intercept (which exponentials, shifted or not, rarely have).

Takeaway: The $y$-intercept requires $x = 0$, not the exponent. Here, $x + 1$ becomes $0 + 1 = 1$ when $x = 0$, giving $2^1 = 2$.

Section 7 — Transformations and Inverses

Transformations modify a graph in predictable ways. Adding a constant outside the function shifts it vertically; adding inside the argument shifts it horizontally — but the direction surprises many students: $f(x - h)$ shifts right (not left). Negating $f(x)$ reflects across the $x$-axis; negating $x$ reflects across the $y$-axis. The inverse of $f$ is found by swapping $x$ and $y$ in the equation and solving for $y$ — it exists (as a function) only when $f$ passes the horizontal line test.

Answer: A

Explanation

Step 1 — Identify the change.

Adding a constant outside the function increases every output by 3, shifting the graph up by 3 units.

Step 2 — Explain errors.

C) Horizontal shift right: A horizontal shift of 3 right would give $y = f(x - 3)$, not $f(x) + 3$.

B) Vertical stretch: A vertical stretch by factor 3 would give $y = 3f(x)$, not $f(x) + 3$.

D) Downward: Adding $+3$ moves up. A downward shift of 3 would be $y = f(x) - 3$.

Takeaway: $y = f(x) + k$: vertical shift UP by $k$ (if $k > 0$). $y = f(x) - k$: vertical shift DOWN.

Answer: D

Explanation

Step 1 — Apply the rule.

Replacing $x$ with $x - h$ shifts the graph right by $h$ units. Here $h = 5$: shift right by 5.

Step 2 — Reason intuitively.

The graph reaches its original position at $x - 5 = 0$, i.e., $x = 5$. The "action" moves to the right.

Step 3 — Explain errors.

A) Left by 5: The most common confusion. $f(x - 5)$ shifts right; $f(x + 5)$ shifts left. The sign inside the argument is opposite to the direction of shift.

B, C) Vertical shifts: Adding or subtracting a constant inside the argument creates a horizontal shift, not a vertical one.

Takeaway: $y = f(x - h)$ shifts right by $h$; $y = f(x + h)$ shifts left by $h$. Remember the counterintuitive rule: the sign inside goes in the opposite direction.

Answer: D

Explanation

Step 1 — Reasoning.

$y = -f(x)$ negates every output: a point $(x, y)$ on the original becomes $(x, -y)$ on the transformed graph. Changing $y$ to $-y$ is the definition of reflection across the $x$-axis.

Step 2 — Explain errors.

B) $y$-axis: Reflecting across the $y$-axis changes $x \to -x$, giving $y = f(-x)$ — the argument is negated, not the output.

C) $y = x$: This gives the inverse function (swap $x$ and $y$).

A) Origin: Reflection through the origin changes $(x, y) \to (-x, -y)$, giving $y = -f(-x)$.

Takeaway: Negate the output ($-f(x)$) → reflect in the $x$-axis. Negate the input ($f(-x)$) → reflect in the $y$-axis.

Answer: A

Explanation

Step 1 — Reasoning.

Replacing $x$ with $-x$ means every point $(x, y)$ becomes $(-x, y)$: the $y$-coordinate is unchanged and the $x$-coordinate is negated. This is reflection across the $y$-axis.

Step 2 — Distinguish from the previous question.

Q49: $-f(x)$ → negate output → $x$-axis reflection. Q50: $f(-x)$ → negate input → $y$-axis reflection.

Why the others are wrong: B ($x$-axis) belongs to $-f(x)$ — the outside negative (see Q49). C (the line $y = x$) is the inverse function's reflection, $f^{-1}$ — a different operation entirely. D (origin) would need both negations at once: $-f(-x)$.

Takeaway: The position of the negative sign tells you which axis: outside the function (output negated) = $x$-axis; inside the argument (input negated) = $y$-axis. Both at once = point reflection through the origin; swap $x$ and $y$ = reflection in $y = x$.

Answer: B

Explanation

Step 1 — Analyse $f(2x)$.

The graph reaches in $x$ what the original reached in $2x$: it gets there "twice as fast" horizontally. Distances along the $x$-axis are halved.

Step 2 — Define compression vs stretch.

A compression by factor 2 means $x$-distances are halved ($\times \frac{1}{2}$). A stretch by factor 2 would double $x$-distances.

Step 3 — Explain errors.

A) Horizontal stretch: That would be $f\!\left(\dfrac{x}{2}\right)$ — dividing inside the argument slows the approach, stretching horizontally.

C) Vertical stretch: A vertical stretch by 2 gives $2f(x)$, not $f(2x)$.

D) Vertical compression: Gives $\frac{1}{2}f(x)$, not $f(2x)$.

Takeaway: Multiplying inside the argument — $f(kx)$ with $k > 1$ — compresses horizontally (divides $x$-scale by $k$). Dividing inside — $f(x/k)$ — stretches horizontally.

Answer: C

Explanation

Step 1 — Replace $f(x)$ with $y$.

$$y = 2x + 5$$

Step 2 — Swap $x$ and $y$.

$$x = 2y + 5$$

Step 3 — Solve for $y$.

$$2y = x - 5 \implies y = \frac{x - 5}{2}$$

$$f^{-1}(x) = \frac{x - 5}{2}$$

Step 4 — Explain errors.

A) $2x - 5$: Student changes the sign of the constant but keeps the coefficient of $x$, applying a "subtract" rather than a proper inversion.

B) $\dfrac{x+5}{2}$: Student swaps the operations but keeps the wrong sign: adds 5 instead of subtracting 5.

D) $5x + 2$: Student swaps the coefficients 2 and 5, a common confusion.

Takeaway: To find an inverse: (1) swap $x$ and $y$, (2) solve for $y$. Each operation in the original is undone in reverse order: $+5$ becomes $-5$; $\times 2$ becomes $\div 2$.

Answer: A

Explanation

Swap and solve: $x = y^2$ gives $y = \pm\sqrt x$.

Choose the branch. The original was restricted to $x \geq 0$, so take the positive root: $y = \sqrt x$. Its domain is the original's range, $x \geq 0$.

Why the others are wrong:

• B (all real $x$) — $\sqrt x$ does not exist for negative $x$.
• C ($x > 0$) excludes 0, but $\sqrt0 = 0$ is valid and matches the original vertex.
• D ($\pm\sqrt x$) is not a function — two outputs per input. The restriction existed precisely to prevent that.

Takeaway: $y=x^2$ has no inverse function until you restrict it — it fails the horizontal line test. Restricting to $x \geq 0$ makes it one-to-one, giving $y = \sqrt x$, positive branch only.

Answer: D

Explanation

Step 1 — Apply the definition of inverse.

By definition, $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$.

$$f(f^{-1}(7)) = 7$$

Step 2 — Explain errors.

A) $f(7)$: Student does not apply $f^{-1}$ at all — treats the composition as just $f$.

C) $1$: No operation on a function composed with its inverse produces 1 in general.

B) $14$: Student doubles the input for no clear reason.

Takeaway: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. A function and its inverse cancel each other out — they are mutual "undoers."

Answer: B

Explanation

Step 1 — Swap $x$ and $y$.

$$x = \frac{3}{y - 1}$$

Step 2 — Solve for $y$.

$$x(y - 1) = 3$$ $$y - 1 = \frac{3}{x}$$ $$y = \frac{3}{x} + 1$$

$$f^{-1}(x) = \frac{3}{x} + 1$$

Step 3 — Explain errors.

A) $\dfrac{3}{1-x}$: Student reflects the denominator sign but does not solve for $y$ properly. The result $\frac{3}{1-x}$ is the same as $\frac{-3}{x-1}$, which is a different function entirely.

C) $\dfrac{x-1}{3}$: Student "cancels" the fraction by moving the denominator to the numerator and dividing — this is the inverse of $y = 3(x-1)$, not $y = \frac{3}{x-1}$.

D) $\dfrac{x}{4}$: Student adds the numerator and denominator: $3 + 1 = 4$. This bears no algebraic relationship to the inverse.

Takeaway: Swap $x$ and $y$, then isolate $y$ through legitimate algebraic steps. Check by verifying $f(f^{-1}(x)) = x$: $f\!\left(\tfrac{3}{x}+1\right) = \frac{3}{\left(\frac{3}{x}+1\right)-1} = \frac{3}{\frac{3}{x}} = x$ ✓.

Section 8 — Interpreting Graphs

The NBT frequently presents a described or sketched graph and asks you to reason about it. Key skills: reading where a function is positive, negative, increasing, or decreasing; identifying zeros and intercepts; recognising asymptotes; and matching an equation to a description of a graph's features.

Answer: C

Explanation

Step 1 — Apply the gradient formula.

$$m = \frac{4 - 0}{0 - (-2)} = \frac{4}{2} = 2$$

Step 2 — Explain errors.

A) $4$: Student reports the $y$-intercept instead of the gradient.

B) $-2$: Student reports the $x$-intercept value instead of the gradient.

D) $\dfrac{1}{2}$: Student inverts the fraction (run over rise instead of rise over run).

Takeaway: Gradient = $\dfrac{\text{rise}}{\text{run}} = \dfrac{\Delta y}{\Delta x}$. Two clean points are given; read them directly and apply the formula without rearranging.

Answer: B

Explanation

Zeros: $x^2-4 = (x-2)(x+2) = 0$, so $x = \pm2$.

Where is it negative? The parabola opens upward, so it dips below the axis between its zeros: $$-2 < x < 2$$

Why the others are wrong:

• A gives where the parabola is positive — outside the zeros.
• C gives the zeros themselves, where $y = 0$, not $y<0$.
• D (all reals) fails at $x=3$, where $y = 5 > 0$.

Takeaway: An upward parabola with zeros $a < b$ is negative between them and positive outside them.

Answer: D

Explanation

Step 1 — Identify the vertex and direction.

Vertex at $(2, 5)$; $a = -1 < 0$ → parabola opens downward.

Step 2 — Determine increasing/decreasing sides.

A downward parabola rises to its vertex and then falls. It is increasing to the left of the vertex ($x < 2$) and decreasing to the right ($x > 2$).

Step 3 — Explain errors.

A) $x > 2$: The function is decreasing after the vertex (past the maximum), not increasing.

C) All real $x$: Parabolas are monotonic on each half, not everywhere.

B) $x = 2$ only: This is where $f$ achieves its maximum — not a meaningful "increasing" interval.

Takeaway: For $a(x-p)^2 + q$: if $a < 0$ (opens down), increasing for $x < p$ and decreasing for $x > p$. Reverse if $a > 0$.

Answer: A

Explanation

Step 1 — Find the zeros.

$(x-1)^2 - 9 = 0 \Rightarrow (x-1)^2 = 9 \Rightarrow x - 1 = \pm 3$.

$$x = 1 + 3 = 4 \quad \text{or} \quad x = 1 - 3 = -2$$

Step 2 — Determine where $f < 0$.

$a = 1 > 0$: upward parabola. Negative between its zeros: $-2 < x < 4$.

Step 3 — Explain errors.

C) Outside interval: The upward parabola is positive outside its zeros, not inside.

B) $-3 < x < 3$: Student uses the zeros of $(x-1)^2 = 9$ as $x = \pm 3$ (ignoring the $-1$ shift).

D) $x > 4$ only: Incomplete — misses the left zero.

Takeaway: First find the zeros; then use the direction of opening to determine the sign in each region. This is the same strategy as solving quadratic inequalities.

Answer: D

Explanation

Step 1 — Find the $x$-coordinate of the vertex.

$$x = -\frac{b}{2a} = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2$$

Step 2 — Evaluate at $x = 2$.

$$y = -2(4) + 8(2) - 5 = -8 + 16 - 5 = 3$$

Step 3 — Explain errors.

A) $y = 5$: Student reports the negative of the constant: $|-{-5}| = 5$.

B) $y = 8$: Student reports $b = 8$ as the maximum value.

C) $y = 11$: Student drops the $-2x^2$ term entirely and evaluates only $8x - 5$ at $x = 2$: $8(2) - 5 = 16 - 5 = 11$. Every term must be included when substituting the vertex $x$-value back in.

Takeaway: Always evaluate the vertex: substitute $x = -\frac{b}{2a}$ back into the original equation. Since $a = -2 < 0$, the vertex is a maximum.

Answer: A

Explanation

Find where they cross: $$x^2 - 4 = x + 2 \implies x^2 - x - 6 = (x-3)(x+2) = 0 \implies x = 3 \text{ or } -2$$

Test between them. At $x=0$: the parabola gives $-4$, the line gives $2$ — so the parabola is below in the middle. It must therefore be above on both outer pieces: $$x < -2 \quad\text{or}\quad x > 3$$

Why the others are wrong:

• B is where the line is above the parabola.
• C and D give only one side. An upward parabola beats the line on both outer regions.

Takeaway: Rewrite "parabola above line" as one inequality, $x^2-x-6 > 0$, then use the rule: an upward parabola is positive outside its zeros. Always confirm with a test value.

Answer: C

Explanation

Step 1 — Check each range.

A) $y = x$: Range is all real numbers (increases without bound in both directions).

C) $y = x^2$: Minimum at $x = 0$ giving $y = 0$; all other outputs are positive. Range: $y \geq 0$. ✓

B) $y = \dfrac{1}{x}$: Takes all real values except 0. Range: $y \neq 0$.

D) $y = x^3$: Takes all real values. Range: all real numbers.

Takeaway: The range $y \geq 0$ is characteristic of even-powered functions like $x^2$, $x^4$, and $|x|$. The squaring operation eliminates negative outputs.

Answer: B

Explanation

Step 1 — Apply the vertical line test.

A) Parabola opening upward: any vertical line crosses it at most once. ✓ (Is a function.)

C) Exponential curve $y = b^x$: strictly increasing or decreasing — any vertical line crosses it once. ✓ (Is a function.)

B) Circle $x^2 + y^2 = r^2$: a vertical line through the interior crosses it at two points (one above and one below the $x$-axis). ✗ (NOT a function.)

D) Logarithmic curve $y = \log_b x$: defined for $x > 0$, strictly monotonic — any vertical line crosses it at most once. ✓ (Is a function.)

Takeaway: The vertical line test is the graphical equivalent of the function definition. A full circle fails it visibly — draw $x = 0$ and it intersects the circle at $(0, r)$ and $(0, -r)$: two outputs for one input.

Answer: D

Explanation

Step 1 — Identify the vertical shift.

The function is in the form $y = \dfrac{a}{x + p} + q$ with $q = -1$.

As $x \to \pm\infty$, $\dfrac{2}{x+3} \to 0$, so $y \to -1$.

Horizontal asymptote: $y = -1$.

Step 2 — Explain errors.

A) $y = 2$: Student reads the numerator as the HA.

B) $y = 3$: Student reads the denominator shift value as the HA (this is the VA: $x = -3$).

C) $y = 0$: Correct for the un-shifted hyperbola $y = \frac{2}{x}$ — ignores the $-1$ translation.

Takeaway: For $y = \frac{a}{x+p} + q$: HA is $y = q$, VA is $x = -p$. The $-1$ outside the fraction shifts the whole graph (including the asymptote) down by 1.

Answer: A

Explanation

Eliminate one feature at a time.

Horizontal asymptote $y=1$ means $q=1$ — that rules out C.

Vertical asymptote $x=-2$ means the denominator is $(x+2)$ — that rules out B, whose $(x-2)$ gives an asymptote at $x=2$.

The $y$-intercept $(0,3)$ settles the last two. Option A: $1 + \frac{4}{2} = 3$ ✓ Option D: $1 - \frac42 = -1$ ✗

Takeaway: For $y = q + \frac{a}{x+p}$: the horizontal asymptote gives $q$, the vertical asymptote gives $p$, and any point on the graph gives $a$. Knock out options one feature at a time rather than solving from scratch.

Section 9 — Absolute-Value Functions (V-Graphs)

The absolute value of a number is its distance from zero, so it is never negative: $|5| = 5$ and $|-5| = 5$. Formally $|x| = \sqrt{x^2}$, which is another way of saying "square it, then take the positive root — the sign is destroyed on the way."

The graph of $y = |x|$ is a V: the right half is the line $y = x$, and the left half is that line reflected upward into $y = -x$. Everything that would have dipped below the $x$-axis is folded back above it.

$$y = a|x - p| + q$$

behaves exactly like the parabola's vertex form: the vertex (the sharp point of the V) sits at $(p, q)$, the V opens upward when $a > 0$ and downward when $a < 0$, and $|a|$ controls how steep the arms are. Two facts do most of the work in exam questions:

• $|\text{anything}| \geq 0$ — always. An absolute value can never equal a negative number.
• $|X| = k$ (with $k > 0$) has two solutions, because $X$ may be $+k$ or $-k$.

Answer: B

Explanation

Compare with $y = a|x-p|+q$: here $p=2$, $q=1$ and $a=+1$. So the vertex is $(2,1)$ and the V opens upward.

The vertex sits where the bracket is zero. Setting $x-2=0$ gives $x=2$, and then $y=0+1=1$.

Check the symmetry: $x=0$ and $x=4$ both give $y=3$ ✓

Why the others are wrong:

• A reads the sign of $p$ straight off the page. $|x-2|$ shifts right by 2, just as $(x-2)^2$ does.
• C flips both the sign of $q$ and the direction. The $+1$ is outside the bars, so it lifts the graph.
• D has the vertex right but the direction wrong. The coefficient is $+1$, so the arms point up.

Takeaway: $y = a|x-p|+q$ has its vertex at $(p,q)$ — read exactly like a parabola's vertex form. The sign of $a$ alone decides up or down.

Answer: C

Explanation

$|X| = 8$ means the inside is 8 units from zero — so it can be $+8$ or $-8$. Split into two ordinary equations:

$$2x - 6 = 8 \quad\Longrightarrow\quad 2x = 14 \quad\Longrightarrow\quad x = 7$$ $$2x - 6 = -8 \quad\Longrightarrow\quad 2x = -2 \quad\Longrightarrow\quad x = -1$$

Check both in the original: $|2(7)-6| = |8| = 8$ ✓ and $|2(-1)-6| = |-8| = 8$ ✓

Why the others are wrong:

• A solves only the positive branch and stops. Every $|X| = k$ with $k>0$ has two answers.
• B deletes the $-6$, solving $|2x| = 8$.
• D takes the second branch but mishandles it — halving before moving the 6, giving $4 - 6 = -2$. Undo the $-6$ first, then divide.

Takeaway: $|X| = k$ (for $k > 0$) always splits into $X = k$ and $X = -k$. Solve both, then substitute both back — the check costs seconds and catches the branch you dropped.

Answer: B

Explanation

$|x|$ is never negative, so $-|x|$ is never positive — it is 0 at $x=0$ and negative everywhere else. Adding 4:

$$y = -|x| + 4 \leq 4$$

The V opens downward with vertex $(0, 4)$, so 4 is the maximum. Test a point: $x=3$ gives $y = 1$, below 4 ✓

Why the others are wrong:

• A treats the vertex as a minimum. The minus sign in front turns the V upside down.
• C ($y \geq 0$) is the range of plain $y=|x|$, ignoring both the reflection and the shift.
• D would need the graph to climb without bound, but nothing here ever exceeds 4.

Takeaway: For $y = a|x-p|+q$: if $a>0$ the range is $y \geq q$; if $a<0$ it is $y \leq q$. Identical to the rule for parabolas.

Answer: A

Explanation

Isolate the absolute value first:

$$|x-3| = 2 - 5 = -3$$

An absolute value is a distance, so it is never negative. No real $x$ works — the solution set is empty.

The graph agrees: $y = |x-3|+5$ is a V with vertex $(3,5)$, so its lowest point is $y=5$. The line $y=2$ passes below it and never meets it.

Why the others are wrong:

• C (two) is the reflex answer. Two is the usual count, but only when the isolated absolute value is positive.
• B (one) happens only when the isolated value is exactly zero — say $|x-3|=0$, giving $x=3$.
• D would need the equation to hold for every $x$.

Takeaway: Isolate the absolute value, then read off the count: a positive right-hand side gives two solutions, zero gives one, negative gives none.

Exam-Bank Extras — Question Types Confirmed in Recent Papers

The NBT MAT reuses question types from a stable bank year after year. The three questions below are modelled directly on types repeatedly confirmed in recent papers and not yet represented in this chapter. Every answer has been independently machine-verified.

Answer: C

Explanation

The recipe never changes: swap $x$ and $y$, then dig $y$ out.

Swap: $\ x = \dfrac{2y}{1-3y}$

Clear the fraction: $\ x(1-3y) = 2y \implies x - 3xy = 2y$

Collect every $y$-term and factor it out — this is the step people miss, because $y$ appears twice: $$x = y(2 + 3x) \implies y = \frac{x}{3x+2}$$

Check: $f(1) = -1$ and $f^{-1}(-1) = 1$ ✓

Why the others are wrong:

• B is the reciprocal $\frac{1}{f(x)}$, not the inverse — the classic confusion.
• D moved $-3xy$ across without flipping its sign.
• A never swapped $x$ and $y$ at all — it rearranged $f$ instead of inverting it.

Takeaway: Swap, clear the fraction, collect all $y$-terms, factor $y$ out, divide. The factoring step is the whole difficulty; the rest is ordinary algebra.

Answer: A

Explanation

To find the original, rewind the transformations — backwards, in reverse order.

Undo the reflection (replace $y$ with $-y$): $$y = -\frac{2}{x-1}$$

Undo "3 left" by shifting 3 right (replace $x$ with $x-3$): $$y = -\frac{2}{(x-3)-1} = -\frac{2}{x-4}$$

Check by running forwards: shift 3 left → $-\frac{2}{x-1}$; reflect → $\frac{2}{x-1}$ ✓

Why the others are wrong:

• C shifted 3 left again instead of reversing.
• B rewound the shift but never the reflection — the minus sign must come back.
• D made both errors at once.

Takeaway: "Before the transformation" means invert each step and reverse their order. Then spend ten seconds running your answer forwards — it turns a risky question into a certain one.

Answer: B

Explanation

A circle at the origin is $x^2+y^2 = r^2$, so the question is really: how far is the nearest point of the hyperbola from the origin?

By symmetry, that nearest point sits on the line $y = x$. Set $y=x$: $$x = \frac{8}{x} \implies x^2 = 8 \implies x = 2\sqrt2$$

The touching point is $(2\sqrt2,\, 2\sqrt2)$, so by Pythagoras: $$r^2 = 8 + 8 = 16$$

Why the others are wrong:

• A ($8$) squared just one coordinate, forgetting to add the second.
• C ($4$) confused $r=4$ with $r^2=4$.
• D ($64$) squared the hyperbola's constant. That 8 is not the radius.

Takeaway: For $y = k/x$, the closest point to the origin always lies on $y=x$ at $x=\sqrt k$, giving the neat general result $x^2+y^2 = 2k$ — here $16$.

Mixed Practice — Chapter 3

These 30 questions mix all Functions and Graphs topics. Work through them without looking back at the sections. Check your answers with the key on the following page — detailed explanations are not given here, so use the earlier sections to diagnose any errors.

Answer: D

Answer: A

Answer: C

Answer: B

Answer: A

Answer: C

Answer: D

Answer: B

Answer: A

Answer: D

Answer: B

Answer: C

Answer: C

Answer: A

Answer: D

Answer: B

Answer: C

Answer: B

Answer: A

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: A

Answer: C

Answer: B

Answer: D

Answer: C

Answer: A