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 $x = 0$.

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