Chapter 4 — Functions

Answer: D

Explanation

Replace every $x$ with $-2\mathclose{,}$ keeping the brackets:

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

Take the power first:

$$(-2)^2 = 4 \quad \text{so} \quad 3 \times 4 = 12$$

Then the middle term. Subtracting a negative adds:

$$-4 \times (-2) = +8$$

$$12 + 8 + 1 = 21$$

Why each wrong option is wrong:

• A, $-19$ — both slips at once: $(-2)^2$ taken as $-4$ and $-4(-2)$ taken as $-8\mathclose{,}$ giving $-12 - 8 + 1\mathclose{.}$
• B, $-3$ — lost the brackets on the square, treating $3(-2)^2$ as $-12\mathclose{,}$ but handled the middle term correctly: $-12 + 8 + 1\mathclose{.}$
• C, $5$ — handled the square correctly but took $-4(-2)$ as $-8\mathclose{.}$

Takeaway: put brackets round a negative input and keep them. The two places signs go wrong are the squared term and the term with a minus in front.

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Answer: G

Explanation

$f(a) = 27$ means the machine produced 27, so set the rule equal to 27 and solve for the input:

$$5a - 8 = 27$$ $$5a = 35$$ $$a = 7$$

Check forwards: $f(7) = 35 - 8 = 27\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $3.8$ — subtracted the 8 rather than adding it.
• H, $35$ — added the 8 correctly and never divided by 5.
• J, $127$ — put 27 into the function instead of solving for the input. Going forwards when the question goes backwards is the standard error here.

Takeaway: $f(a) = 27$ gives you the output. Set the rule equal to it and solve. Then substitute your answer forwards to check.

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Answer: C

Explanation

Work from the inside out. The inner bracket is $g(3)\mathclose{:}$

$$g(3) = 3^2 = 9$$

Now that 9 is the input to $f\mathclose{:}$

$$f(9) = 9 - 4 = 5$$

Why each wrong option is wrong:

• A, $-1$ — worked out $f(3)$ and stopped.
• B, $1$ — applied $f$ first and then $g\mathclose{:}$ $f(3) = -1\mathclose{,}$ and $(-1)^2 = 1\mathclose{.}$ Order matters, and $f(g(x))$ is generally not $g(f(x))\mathclose{.}$
• D, $13$ — added 4 instead of subtracting.

Takeaway: in $f(g(x))\mathclose{,}$ $g$ goes first — it is the one wrapped in the inner brackets. Work outwards from the innermost bracket, exactly as you would with arithmetic.

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Answer: G

Explanation

Undo the outer function first. Whatever $g(a)$ is, $f$ turned it into 33:

$$2 \cdot g(a) + 1 = 33 \quad\Longrightarrow\quad 2 \cdot g(a) = 32 \quad\Longrightarrow\quad g(a) = 16$$

Now undo the inner one:

$$a^2 = 16 \qquad a = \pm 4$$

The question asks for the positive input, so $a = 4\mathclose{.}$

Check forwards, which is always worth doing on a composite: $g(4) = 16\mathclose{,}$ and $f(16) = 32 + 1 = 33\mathclose{.}$ Correct.

Why each wrong option is wrong:

• F, $2$ — took the square root twice, $\sqrt{\sqrt{16}} = 2\mathclose{.}$
• H, $8$ — halved 16 instead of taking its square root.
• J, $16$ — gave $g(a) = 16\mathclose{,}$ the value halfway through the chain, rather than $a$ itself. Stopping one function early is the standard error here.

Takeaway: peel a composite from the outside in — undo $f\mathclose{,}$ then undo $g$ — which is the reverse of the order you would use to evaluate it. Then run your answer forwards through both functions to check.

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Section 2 — Graphs, Key Features and Transformations

Reading a graph is mostly vocabulary:

• an $x$-intercept is where the graph crosses the $x$-axis, so $y = 0\mathclose{.}$ It is also called a zero or a root.
• a $y$-intercept is where it crosses the $y$-axis, so $x = 0\mathclose{.}$
• a vertex is the turning point of a parabola.

Transformations follow one rule that catches everybody: changes inside the brackets act on $x$ and go the opposite way to what they look like; changes outside act on $y$ and behave normally.

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Answer: B

Explanation

The change is inside the brackets, so it moves the graph horizontally — and inside-the-bracket changes go the opposite way to their sign. A minus 3 shifts the graph 3 units to the right.

$$(2, 5) \longrightarrow (2 + 3, \; 5) = (5, 5)$$

If the direction feels backwards, reason it through: the new graph at $x = 5$ computes $f(5 - 3) = f(2) = 5\mathclose{.}$ So the height that used to be at $x = 2$ now appears at $x = 5$ — three units to the right.

Why each wrong option is wrong:

• A, $\left( -1, \ 5\right)$ — moved left, the direction the minus sign suggests.
• C, $\left( 2, \ 8\right)$ and D, $\left( 2, \ 2\right)$ — changed the $y$-coordinate. A change inside the brackets never moves a graph vertically.

Takeaway: inside the brackets means horizontal and opposite; outside means vertical and normal. When in doubt, substitute one $x$ and see which output comes back.

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Answer: G

Explanation

The $+6$ is outside the brackets, so it moves the graph vertically, upwards, by 6. The $x$-coordinate of the minimum does not change — the graph slides straight up.

$$(1, -4) \longrightarrow (1, \; -4 + 6) = (1, 2)$$

Why each wrong option is wrong:

• H, $\left( 7, \ -4\right)$ and J, $\left( -5, \ -4\right)$ — moved horizontally. Outside the brackets is always vertical.
• F, $\left( 1, \ -10\right)$ — moved down.

Takeaway: outside the brackets, the change acts on the output, so only the $y$-coordinate moves — and it moves the way the sign says.

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Answer: D

Explanation

Take the two shifts one at a time.

Right by 4 is a horizontal shift, so it goes inside the brackets, and inside changes act opposite to their sign. Right means minus:

$$f(x - 4)$$

Down by 2 is vertical, so it goes outside, and outside changes behave normally. Down means minus:

$$f(x - 4) - 2$$

If the horizontal rule still feels wrong, test it on $f(x) = x^2\mathclose{,}$ whose lowest point is at $x = 0\mathclose{.}$ The graph of $(x-4)^2$ has its lowest point where $x - 4 = 0\mathclose{,}$ that is at $x = 4$ — four units to the right. Confirmed.

Why each wrong option is wrong:

• B, $y = f(x + 4) - 2$ — wrote the right-shift as $+4\mathclose{.}$
• C, $y = f(x - 4) + 2$ — wrote the downward shift as $+2\mathclose{.}$
• A, $y = f(x + 4) + 2$ — both signs the wrong way round.

Takeaway: horizontal shifts are inside and opposite; vertical shifts are outside and obvious. Do them one at a time, and check with $x^2$ if unsure.

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Answer: J

Explanation

A fraction is undefined only when its bottom is zero:

$$x - 5 = 0 \qquad x = 5$$

At $x = 5$ the function reads $\frac{8}{0}\mathclose{,}$ which has no value.

Why each wrong option is wrong:

• F, $-3$ — set the numerator to zero. That gives a zero of the function, $f(-3) = 0\mathclose{,}$ which is a perfectly good value, not an excluded one.
• G, $0$ — $f(0) = \frac{3}{-5}\mathclose{,}$ which is fine.
• H, $3$ — took a number from the top.

Takeaway: only the denominator can make a function undefined. Setting the numerator to zero answers a different question — where the graph crosses the $x$-axis.

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Section 3 — Logarithms

A logarithm asks: what power do I need?

$$\log_b(y) = x \quad \text{means exactly} \quad b^{x} = y$$

Read $\log_2(8) = 3$ as "2 to the power 3 is 8". Converting between those two forms answers most ACT logarithm questions on its own.

The three rules worth knowing:

$$\log(mn) = \log m + \log n$$ $$\log\!\left(\frac{m}{n}\right) = \log m - \log n$$ $$\log(m^k) = k \log m$$

Multiplying inside becomes adding outside — which is what logarithms were invented for.

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Answer: D

Explanation

$\log_3(x) = 4$ says: 3 to the power 4 is $x\mathclose{.}$

$$x = 3^4 = 81$$

The pattern to hold on to: the base goes to the bottom of the power, and the answer of the log becomes the exponent.

Why each wrong option is wrong:

• A, $\frac{4}{3}$ — divided.
• B, $12$ — multiplied 3 by 4.
• C, $64$ — computed $4^3 = 64\mathclose{,}$ swapping base and exponent. The two are not interchangeable.

Takeaway: $\log_b(y) = x$ means $b^x = y\mathclose{.}$ Write the exponential form out every time; it turns a log question into an ordinary powers question.

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Answer: J

Explanation

Two routes, both quick.

Directly: $\log_2(8) = 3$ because $2^3 = 8\mathclose{,}$ and $\log_2(4) = 2$ because $2^2 = 4\mathclose{.}$ So the sum is $3 + 2 = 5\mathclose{.}$

By the rule: adding logs multiplies their insides:

$$\log_2(8) + \log_2(4) = \log_2(32) = 5$$

because $2^5 = 32\mathclose{.}$ Same answer, which is a good check that the rule is being used correctly.

Why each wrong option is wrong:

• F, $32$ — multiplied the insides correctly to get 32, then forgot to take the logarithm of it.
• G, $12$ — added the insides. Adding logs multiplies the insides, it does not add them.
• H, $6$ — multiplied the two logarithms, $3 \times 2\mathclose{,}$ instead of adding them.

Takeaway: adding logs multiplies what is inside. When the numbers are small powers of the base, evaluating each log separately is faster than using the rule — and doing both is a free check.

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Section 4 — Modelling with Functions

The ACT writes growth and decay in function notation, and the shape to recognise is:

$$f(x) = a \cdot b^{\,x}$$

• $a$ is the starting value, what you have when $x = 0$
• $b$ is the multiplier per step: above 1 is growth, below 1 is decay

A 15% rise per year means $b = 1.15\mathclose{.}$ A 12% fall means $b = 0.88$ — and note it is $0.88\mathclose{,}$ not $-0.12\mathclose{.}$

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Answer: D

Explanation

Growth of 15% per hour multiplies by $1.15$ each hour — the original whole plus another 15%:

$$P = 400(1.15)^{h}$$

Check one hour: $400 \times 1.15 = 460\mathclose{,}$ which is $400 + 60\mathclose{,}$ and 60 is 15% of 400. Correct.

Why each wrong option is wrong:

• B, $P = 400(0.15)^h$ — used $0.15$ as the multiplier. That would leave 15% of the colony each hour, an 85% collapse, not growth.
• C, $P = 400 + 15h$ — modelled it as linear, adding 15 bacteria per hour. A percentage applies to the current amount, so the increase grows as the colony does.
• A, $P = 400(15)^h$ — used 15, multiplying the colony fifteenfold every hour.

Takeaway: a $p\%$ rise per period is a multiplier of $1 + \frac{p}{100}\mathclose{;}$ a $p\%$ fall is $1 - \frac{p}{100}\mathclose{.}$ Always test one period against the plain arithmetic.

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Answer: F

Explanation

A multiplier of $0.88$ means the car keeps 88% of its value each year. What it loses is the rest:

$$100\% - 88\% = 12\%$$

Check with numbers. After one year:

$$24000 \times 0.88 = 21120$$

a fall of 2880, and $\frac{2880}{24000} = 0.12 = 12\%\mathclose{.}$ Correct.

Why each wrong option is wrong:

• H, The car loses 88% of its value each year — read $0.88$ as the amount lost rather than the amount kept. Losing 88% a year would leave the car worth 2880 after twelve months.
• G, The car loses $88 each year — read it as a fixed number of dollars. Exponential models work in proportions, so the dollar loss shrinks each year.
• J, The car gains 88% of its value each year — a multiplier below 1 always shrinks a quantity.

Takeaway: the multiplier is what is kept. Subtract it from 1 to find what is lost. Below 1 is decay; above 1 is growth.

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Answer: C

Explanation

Count the doubling periods, not the years:

$$\frac{20 \text{ years}}{5 \text{ years per doubling}} = 4 \text{ doublings}$$

$$800 \times 2^4 = 800 \times 16 = 12800$$

Written as a formula, that is $P = 800 \cdot 2^{t/5}$ — the division by 5 is what converts years into periods.

Check by doubling four times: 800, 1600, 3200, 6400, 12800. Correct.

Why each wrong option is wrong:

• A, $1600$ — doubled once, as though the whole 20 years were a single period.
• B, $3200$ — multiplied by 4, the number of doublings, instead of doubling four times. Four doublings multiply by $2^4 = 16\mathclose{.}$
• D, $838860800$ — doubled every year, giving a number in the hundreds of millions.

Takeaway: when the growth period is not 1, the exponent is $\frac{\text{time}}{\text{period}}\mathclose{.}$ Count the periods first, then apply the multiplier that many times.

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Answer: F

Explanation

Let $p$ be the starting count. Four days of tripling multiplies it by $3^4\mathclose{:}$

$$p \times 3^4 = 810$$ $$81p = 810$$ $$p = 10$$

Check forwards: 10, 30, 90, 270, 810. Four triplings. Correct.

Why each wrong option is wrong:

• G, $30$ — divided by $3^3\mathclose{,}$ undoing three days where four were needed.
• H, $270$ — divided by 3 once, undoing a single day.
• J, $2430$ — multiplied, growing the colony further into the future.

Takeaway: to go backwards through exponential growth, divide by the multiplier raised to the number of periods. Then run it forwards to check — five seconds, and it catches every off-by-one.

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Answer: D

Explanation

Work from the inside out, reading the table instead of a formula.

First $g(2)\mathclose{:}$ in the $g$ row, under $x = 2\mathclose{,}$ is 3.

That 3 is now the input to $f\mathclose{.}$ In the $f$ row, under $x = 3\mathclose{,}$ is 4:

$$f(g(2)) = f(3) = 4$$

Why each wrong option is wrong:

• A, $1$ — read $f(2)$ straight off the table, ignoring $g$ altogether.
• B, $2$ — went the other way: $f(2) = 1\mathclose{,}$ then $g(1) = 2\mathclose{.}$ That is $g(f(2))\mathclose{,}$ a different question.
• C, $3$ — found $g(2) = 3$ and stopped.

Takeaway: a composite works from a table exactly as it does from a formula. The inner function goes first, and its output is the next input.

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Answer: J

Explanation

Substitute the whole of $g(x)$ into $f\mathclose{:}$

$$f(g(x)) = (x - 3)^2 = x^2 - 6x + 9$$

Check at $x = 5\mathclose{:}$ $g(5) = 2$ and $f(2) = 4\mathclose{;}$ and $25 - 30 + 9 = 4\mathclose{.}$ ✓

Why each wrong option is wrong:

• F, $x^{2} - 3$ — computed $g(f(x)) = x^2 - 3\mathclose{,}$ the other order.
• G, $x^{2} - 9$ — squared each term, forgetting the middle term.
• H, $x^{2} + 6 x + 9$ — used $(x + 3)^2\mathclose{.}$

Takeaway: substitute the inner function whole, brackets and all, then expand. A squared bracket always has a middle term.

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Answer: D

Explanation

Write $y = 3x + 4\mathclose{,}$ swap the letters, and solve for the new $y\mathclose{:}$

$$x = 3y + 4 \quad\Longrightarrow\quad y = \frac{x - 4}{3}$$

Check: $f(2) = 10\mathclose{,}$ and $f^{-1}(10) = \frac{6}{3} = 2\mathclose{.}$ ✓

Why each wrong option is wrong:

• B, $\frac{x + 4}{3}$ — added rather than subtracted.
• C, $3 x - 4$ — reversed the signs without reversing the operations.
• A, $\frac{1}{3 x + 4}$ — an inverse function is not a reciprocal. The notation $f^{-1}$ does not mean "one over".

Takeaway: swap $x$ and $y\mathclose{,}$ then solve. $f^{-1}$ undoes $f\mathclose{;}$ it is not $\frac{1}{f}\mathclose{.}$

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Answer: H

Explanation

$$f(3) = 6 - 1 = 5 \qquad f(5) = 10 - 1 = 9$$

Why each wrong option is wrong:

• F, $5$ — applied $f$ once.
• G, $6$ — doubled and never subtracted.
• J, $11$ — used 6 rather than 5 as the second input, dropping the first subtraction.

Takeaway: $f(f(x))$ means run the machine twice. Write the intermediate value down before feeding it back in.

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Answer: B

Explanation

The expression under a square root cannot be negative, so

$$x - 5 \ge 0 \quad\Longrightarrow\quad x \ge 5$$

The end is included: at $x = 5$ the root is $\sqrt{0} = 0\mathclose{,}$ a perfectly good value.

Why each wrong option is wrong:

• A, $x > 5$ — excluded $x = 5\mathclose{.}$ Zero under a root is fine; only negatives are not.
• C, $x \le 5$ — reversed the inequality, which is the region where the inside is negative.
• D, $x \ge -5$ — a sign error on the boundary.

Takeaway: under a square root, set the inside $\ge 0\mathclose{.}$ Note it is greater than or equal, because $\sqrt{0}$ exists.

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Answer: H

Explanation

The input is $5\mathclose{,}$ and $5 \ge 3\mathclose{,}$ so the second rule applies:

$$f(5) = 5^2 - 4 = 25 - 4 = 21$$

The first rule is only for inputs below 3, and it plays no part here.

Why each wrong option is wrong:

• F, $5$ — chose the right rule and then put in 3, the boundary, instead of 5, the input.
• G, $11$ — used the rule for $x < 3\mathclose{,}$ which does not apply to 5.
• J, $25$ — used the right rule and dropped the $-4\mathclose{.}$

Takeaway: with a piecewise function, check the condition first and pick the line, then substitute. The other line is irrelevant.

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Answer: B

Explanation

Check the condition before choosing a line. The first says $x \le 10\mathclose{,}$ and 10 satisfies it, so:

$$c(10) = 5 + 2(10) = 25$$

The second line applies only above 10. Note the two rules do not agree at the boundary — $c$ jumps from 25 to 30 — which is why the $\le$ matters.

Why each wrong option is wrong:

• A, $15$ — dropped the coefficient of $x\mathclose{.}$
• C, $30$ — used the rule for $x > 10\mathclose{;}$ 10 is not greater than 10.
• D, $35$ — added both rules together.

Takeaway: read the condition first and note whether it uses $<$ or $\le\mathclose{.}$ That single symbol decides which rule owns the boundary.

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Answer: F

Explanation

$x^2$ is at its smallest when $x = 0\mathclose{,}$ where it equals 0. Everywhere else it is positive. So

$$f(x) = x^2 + 3 \ge 0 + 3 = 3$$

The value 3 is reached, at $x = 0\mathclose{,}$ so the end is included.

Why each wrong option is wrong:

• G, $y > 3$ — excluded 3, but $f(0) = 3$ exactly.
• H, $y \le 3$ — reversed the inequality; the parabola opens upward, so 3 is a minimum, not a maximum.
• J, $y \ge 0$ — gave the range of $x^2$ alone, forgetting the $+3$ lifts every output.

Takeaway: for range questions, find the extreme value the expression can reach and ask whether it is actually attained. A square's minimum is 0, at $x = 0\mathclose{.}$

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Answer: B

Explanation

$|x - 4|$ is smallest when the inside is zero, which happens at $x = 4\mathclose{.}$ There it equals 0, so

$$f(4) = 0 + 1 = 1$$

Everywhere else $|x - 4|$ is positive, so $f$ is larger.

Why each wrong option is wrong:

• A, $0$ — gave the minimum of $|x-4|$ and forgot the $+1\mathclose{.}$
• C, $4$ — gave $x = 4\mathclose{,}$ the input where the minimum occurs, not the minimum itself.
• D, $5$ — added the two constants.

Takeaway: an absolute value bottoms out at zero where its inside is zero. Then read whether the question wants the input or the output.

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Answer: F

Explanation

Ask: 2 to what power gives 64?

$$2^6 = 64 \quad\Longrightarrow\quad \log_2(64) = 6$$

Counting the doublings is often quicker than any rule: 2, 4, 8, 16, 32, 64 — six of them.

Why each wrong option is wrong:

• G, $8$ — took a square root.
• H, $32$ — divided 64 by 2. A logarithm is an exponent, not a quotient.
• J, $128$ — multiplied by the base.

Takeaway: $\log_b(y)$ asks "$b$ to what power gives $y\mathclose{?}$" With small numbers, count the multiplications.

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Answer: B

Explanation

Ask: 9 to what power gives 3?

$$9^{1/2} = \sqrt{9} = 3 \quad\Longrightarrow\quad \log_9(3) = \tfrac{1}{2}$$

The check that settles it instantly: whenever the number is smaller than the base, the logarithm is between 0 and 1.

Why each wrong option is wrong:

• A, $\frac{1}{3}$ — divided 3 by 9. A logarithm is an exponent, not a quotient.
• C, $2$ — computed $\log_3(9) = 2\mathclose{,}$ the two arguments swapped. Note it is the reciprocal of the right answer, which is no coincidence.
• D, $3$ — divided 9 by 3.

Takeaway: a logarithm is between 0 and 1 exactly when the number is between 1 and the base. Use that to sanity-check before calculating.

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Answer: F

Explanation

Subtracting logs of the same base divides their insides:

$$\log_5(50) - \log_5(2) = \log_5\!\left(\frac{50}{2}\right) = \log_5(25)$$

And $5^2 = 25\mathclose{,}$ so the answer is 2.

Why each wrong option is wrong:

• G, $25$ — divided correctly to reach 25 and then forgot to take the logarithm.
• H, $48$ — subtracted the insides. Subtracting logs divides; it does not subtract.
• J, $100$ — multiplied, which is what adding logs does.

Takeaway: adding logs multiplies inside; subtracting divides. Then finish by evaluating the log you are left with.

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Answer: A

Explanation

The fixed charge is paid once, so remove it before dividing:

$$4 + 2.5m = 29$$ $$2.5m = 25 \qquad m = 10$$

Check: $2.5 \times 10 = 25\mathclose{,}$ and $25 + 4 = 29\mathclose{.}$ ✓

Why each wrong option is wrong:

• B, $11.6$ — divided the whole fare by the rate, as though there were no fixed charge. That is the trap this question is built around.
• C, $13.2$ — added the fixed charge instead of subtracting it.
• D, $25$ — took off the fixed charge and stopped. 25 is the dollars charged for distance, not the distance.

Takeaway: in $C = \text{fixed} + \text{rate} \times x\mathclose{,}$ subtract the fixed part before dividing by the rate.

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Answer: H

Explanation

The halving period is 4 days, not 1, so the exponent counts periods:

$$\text{periods} = \frac{d}{4} \qquad\Longrightarrow\qquad V = 200(0.5)^{d/4}$$

Check 4 days: the exponent is 1, giving $200 \times 0.5 = 100\mathclose{.}$ ✓ Check 8 days: the exponent is 2, giving 50. ✓

Why each wrong option is wrong:

• F, $200(0.5)^{d}$ — halves every day, so by day 4 it would be 12.5.
• G, $200 - 25d$ — linear decay, which would reach zero on day 8 and go negative after.
• J, $200(4)^{d/2}$ — grows the quantity instead of shrinking it.

Takeaway: when the period is not 1, the exponent is $\frac{\text{time}}{\text{period}}\mathclose{.}$ Test the model at exactly one period.

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Answer: C

Explanation

The study began at $t = 0\mathclose{,}$ and anything to the power 0 is 1:

$$P(0) = 1200 \times (1.03)^0 = 1200 \times 1 = 1200$$

So the number in front of the bracket is always the starting value.

Why each wrong option is wrong:

• A, $0$ — a model does not have to start at zero; this one starts at 1200.
• B, $1.03$ — gave the multiplier, which describes growth rather than size.
• D, $1236$ — gave $P(1)\mathclose{,}$ the population after one year.

Takeaway: in $a \cdot b^t\mathclose{,}$ $a$ is where it starts and $b$ is how it changes. Setting $t = 0$ recovers $a$ every time, because $b^0 = 1\mathclose{.}$

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Answer: F

Explanation

$$f\!\left(\tfrac{1}{4}\right) = 8\left(\tfrac{1}{4}\right) - 3 = 2 - 3 = -1$$

Only the $x$ is replaced. The $-3$ is unaffected by the input.

Why each wrong option is wrong:

• G, $1.25$ — divided the 3 by 4 as well, but the 3 is not multiplied by $x\mathclose{.}$
• H, $2$ — multiplied correctly and stopped before subtracting.
• J, $5$ — ignored the input and used $x = 1\mathclose{.}$

Takeaway: substitute for every $x$ and nothing else. Terms without an $x$ stay exactly as they are.

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Answer: C

Explanation

Read the two outputs off the table and subtract:

$$f(3) = 11 \qquad f(1) = 5 \qquad f(3) - f(1) = 6$$

Why each wrong option is wrong:

• A, $2$ — subtracted the inputs, $3 - 1\mathclose{,}$ rather than the outputs.
• B, $3$ — gave the step between consecutive rows, which is the slope of this linear function but not what was asked.
• D, $16$ — added the two outputs.

Takeaway: $f(3)$ means the output when the input is 3 — look it up in the bottom row. The top row holds inputs, never answers.

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Answer: J

Explanation

Work from the inner bracket outwards:

$$f(4) = 6 \qquad g(6) = 18$$

Why each wrong option is wrong:

• F, $6$ and G, $12$ — stopped after one function.
• H, $14$ — applied $g$ first: $g(4) = 12\mathclose{,}$ then $f(12) = 14\mathclose{.}$ The order is not interchangeable, and this question is the mirror image of one where it is the other way round.

Takeaway: the function named inside the brackets goes first. In $g(f(x))$ that is $f\mathclose{;}$ in $f(g(x))$ it is $g\mathclose{.}$

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Answer: A

Explanation

$$f(5) = 2(5) + 6 = 16 \qquad g(16) = \frac{16 - 6}{2} = 5$$

The answer is the number we started with — and that is no accident. $g$ undoes exactly what $f$ does, so they are inverse functions, and $g(f(x)) = x$ for every $x\mathclose{.}$

Why each wrong option is wrong:

• B, $8$ — halved 16 without subtracting the 6 first.
• C, $10$ — subtracted the 6 and forgot to halve.
• D, $16$ — gave $f(5)$ and stopped halfway.

Takeaway: if a composite hands back the input you started with, the two functions are inverses. Spotting that saves the arithmetic entirely.

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Answer: J

Explanation

Set $x = 0\mathclose{:}$

$$f(0) = 0 - 0 + 7 = 7$$

So the constant term is the $y$-intercept, for any polynomial — every other term contains an $x$ and vanishes at zero.

Why each wrong option is wrong:

• F, $-4$ — gave the coefficient of $x\mathclose{.}$
• G, $0$ — a graph need not pass through the origin, and this one does not.
• H, $3$ — the lowest value of the function, reached at the vertex $x = 2\mathclose{.}$ That is where the graph turns, not where it meets the $y$-axis.

Takeaway: the $y$-intercept is the value at $x = 0\mathclose{,}$ which for a polynomial is just the constant term.

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