Secondary 3 Elementary Mathematics Algebra Functions Quiz

Secondary 3 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

Total Marks: 40

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Section A (Questions 1–5, 1 mark each)

1. Given $f(x) = 3x^2 - 4x + 5$, find $f(-2)$.

Working: $f(-2) = 3(-2)^2 - 4(-2) + 5$ $= 3(4) + 8 + 5$ $= 12 + 8 + 5$ $= 25$

Answer: 25 [1]

Marking note: Award 1 mark for correct final answer. No working required for 1-mark question, but substitution must be correct if shown.

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2. The function $g(x) = \frac{2x - 1}{x + 3}$ for $x \neq -3$. State the value that $x$ cannot take.

Explanation: The denominator cannot be zero. $x + 3 = 0 \Rightarrow x = -3$.

Answer: $x \neq -3$ or $x = -3$ [1]

Marking note: Accept "$-3$" or "$x \neq -3$". The question asks for "the value that $x$ cannot take", so "$-3$" is the direct answer.

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3. The graph of $y = x^2 - 6x + 8$ cuts the $x$-axis at points $A$ and $B$. Write down the coordinates of $A$ and $B$.

Working: $x^2 - 6x + 8 = 0$ $(x - 2)(x - 4) = 0$ $x = 2$ or $x = 4$

Answer: $A(2, 0)$, $B(4, 0)$ [1] (order does not matter)

Marking note: 1 mark for both correct coordinates. Must be written as coordinate pairs $(x, 0)$.

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4. $h(x) = 5 - 2x$. Find $x$ when $h(x) = 11$.

Working: $5 - 2x = 11$ $-2x = 6$ $x = -3$

Answer: $-3$ [1]

Marking note: 1 mark for correct answer. Accept $x = -3$.

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5. From the graph of $y = f(x)$ with vertex $(0, 4)$ and domain $-3 \le x \le 3$.

Explanation: The vertex $(0, 4)$ is the maximum point (parabola opens downwards). The minimum occurs at the endpoints $x = \pm 3$. By symmetry, $f(-3) = f(3)$. From the graph, the $x$-intercepts are at $(-2, 0)$ and $(2, 0)$, so the function goes down to 0. The range is from the minimum value (0) to the maximum value (4).

Answer: $0 \le f(x) \le 4$ or $[0, 4]$ [1]

Marking note: 1 mark for correct range. Must use correct inequality notation or interval notation. Accept $0 \le y \le 4$.

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Section B (Questions 6–15, 2 marks each)

6. $f(x) = 2x^2 - 8x + 7$

(a) Express in form $a(x - h)^2 + k$.

Working: $f(x) = 2(x^2 - 4x) + 7$ $= 2[(x - 2)^2 - 4] + 7$ $= 2(x - 2)^2 - 8 + 7$ $= 2(x - 2)^2 - 1$

Answer: $2(x - 2)^2 - 1$ [1]

(b) State minimum value and $x$ at which it occurs.

Explanation: In vertex form $a(x - h)^2 + k$ with $a > 0$, minimum value is $k$ at $x = h$. Here $a = 2 > 0$, $h = 2$, $k = -1$.

Answer: Minimum value = $-1$, at $x = 2$ [1]

Marking note: 1 mark each part. For (a), must show completing the square correctly. For (b), follow-through from (a) allowed if vertex form is correct.

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7. $g(x) = \frac{3}{x - 2} + 1$, $x \neq 2$

(a) Vertical and horizontal asymptotes.

Explanation: Vertical asymptote where denominator is zero: $x - 2 = 0 \Rightarrow x = 2$. Horizontal asymptote: as $x \to \pm\infty$, $\frac{3}{x-2} \to 0$, so $y \to 1$.

Answer: Vertical asymptote: $x = 2$, Horizontal asymptote: $y = 1$ [1]

(b) $y$-intercept coordinates.

Working: Set $x = 0$: $g(0) = \frac{3}{0 - 2} + 1 = -\frac{3}{2} + 1 = -\frac{1}{2}$.

Answer: $(0, -\frac{1}{2})$ or $(0, -0.5)$ [1]

Marking note: 1 mark each part. For (b), must give coordinates, not just $y$-value.

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8. $h(x) = x^2 - 4x - 5$ for $x \ge 2$

(a) Explain why $h$ has an inverse.

Explanation: The function $h(x) = (x - 2)^2 - 9$ has vertex at $(2, -9)$. For $x \ge 2$, the function is strictly increasing (right side of vertex). A strictly monotonic function is one-to-one, hence has an inverse.

Answer: $h$ is strictly increasing on $x \ge 2$ (or one-to-one on this domain) [1]

(b) Find $h^{-1}(x)$ and its domain.

Working: Let $y = x^2 - 4x - 5 = (x - 2)^2 - 9$ for $x \ge 2$. Swap $x$ and $y$: $x = (y - 2)^2 - 9$ $(y - 2)^2 = x + 9$ $y - 2 = \sqrt{x + 9}$ (positive root since $y \ge 2$) $y = 2 + \sqrt{x + 9}$

Domain of $h^{-1}$ = Range of $h$ = $[-9, \infty)$

Answer: $h^{-1}(x) = 2 + \sqrt{x + 9}$, Domain: $x \ge -9$ [1]

Marking note: 1 mark for correct inverse expression with correct root choice, 1 mark for correct domain. Must show positive square root chosen because $x \ge 2$ in original.

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9. $y = \frac{k}{x}$ passes through $(2, 6)$

(a) Find $k$.

Working: $6 = \frac{k}{2} \Rightarrow k = 12$

Answer: $12$ [1]

(b) Find $y$ when $x = 4$.

Working: $y = \frac{12}{4} = 3$

Answer: $3$ [1]

Marking note: 1 mark each. Part (b) follow-through from (a) allowed.

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10. $f(x) = 2x + 3$, $g(x) = x^2 - 1$

(a) Find $fg(2)$.

Working: $fg(2) = f(g(2)) = f(2^2 - 1) = f(3) = 2(3) + 3 = 9$

Answer: $9$ [1]

(b) Solve $gf(x) = 15$.

Working: $gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 - 1 = 15$ $(2x + 3)^2 = 16$ $2x + 3 = \pm 4$ $2x = 1$ or $2x = -7$ $x = 0.5$ or $x = -3.5$

Answer: $x = 0.5$ or $x = -3.5$ [1]

Marking note: 1 mark each. For (b), both solutions required for the mark. Must consider both $\pm$ square roots.

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11. $f(x) = -2x^2 + 12x - 13$

(a) Express in form $a(x - h)^2 + k$.

Working: $f(x) = -2(x^2 - 6x) - 13$ $= -2[(x - 3)^2 - 9] - 13$ $= -2(x - 3)^2 + 18 - 13$ $= -2(x - 3)^2 + 5$

Answer: $-2(x - 3)^2 + 5$ [1]

(b) Sketch graph for $-1 \le x \le 7$.

Key features:

• Vertex: $(3, 5)$ (maximum, since $a = -2 < 0$)
• $y$-intercept: $x = 0 \Rightarrow f(0) = -13$, so $(0, -13)$
• $x$-intercepts: Solve $-2x^2 + 12x - 13 = 0 \Rightarrow 2x^2 - 12x + 13 = 0$ $x = \frac{12 \pm \sqrt{144 - 104}}{4} = \frac{12 \pm \sqrt{40}}{4} = 3 \pm \frac{\sqrt{10}}{2} \approx 1.42, 4.58$ (both in domain)
• Endpoints: $f(-1) = -2 - 12 - 13 = -27$, $f(7) = -98 + 84 - 13 = -27$

Answer: See sketch [2]

Marking note: 2 marks for sketch:

• 1 mark: Correct shape (downward parabola), vertex at $(3, 5)$ labelled, $y$-intercept at $(0, -13)$ labelled
• 1 mark: Correct $x$-intercepts shown (approx), endpoints at $x = -1$ and $x = 7$ shown, axes scaled appropriately

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12. $f(x) = \sqrt{x + 4}$ for $x \ge -4$

(a) Find $f(5)$.

Working: $f(5) = \sqrt{5 + 4} = \sqrt{9} = 3$

Answer: $3$ [1]

(b) Find $x$ when $f(x) = 5$.

Working: $\sqrt{x + 4} = 5 \Rightarrow x + 4 = 25 \Rightarrow x = 21$

Answer: $21$ [1]

Marking note: 1 mark each. For (b), must square both sides correctly and check $x \ge -4$ (21 satisfies this).

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13. $f(x) = ax^2 + bx + c$, max at $(2, 9)$, passes through $(0, 5)$.

Working: Vertex form: $f(x) = a(x - 2)^2 + 9$ Passes through $(0, 5)$: $5 = a(0 - 2)^2 + 9 = 4a + 9$ $4a = -4 \Rightarrow a = -1$ $f(x) = -(x - 2)^2 + 9 = -(x^2 - 4x + 4) + 9 = -x^2 + 4x + 5$ So $a = -1$, $b = 4$, $c = 5$

Answer: $a = -1$, $b = 4$, $c = 5$ [2]

Marking note: 2 marks: 1 for $a = -1$, 1 for $b = 4$ and $c = 5$ (both correct). Alternative method: use vertex formula $x = -\frac{b}{2a} = 2$ and $f(2) = 9$, $f(0) = 5 = c$.

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14. $f(x) = 3x - 2$, $g(x) = \frac{x + 2}{3}$

(a) Show $gf(x) = x$.

Working: $gf(x) = g(f(x)) = g(3x - 2) = \frac{(3x - 2) + 2}{3} = \frac{3x}{3} = x$

Answer: Shown [1]

(b) Relationship between $f$ and $g$.

Answer: $f$ and $g$ are inverse functions of each other (or $g = f^{-1}$ and $f = g^{-1}$) [1]

Marking note: 1 mark each. For (a), must show clear substitution and simplification. For (b), "inverse functions" is the key phrase.

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15. $h(x) = x^3 - 3x^2 + 2$

(a) Find $h(-1)$.

Working: $h(-1) = (-1)^3 - 3(-1)^2 + 2 = -1 - 3 + 2 = -2$

Answer: $-2$ [1]

(b) Factorise completely given $x = 1$ is a root.

Working: Since $x = 1$ is a root, $(x - 1)$ is a factor. Divide: $(x^3 - 3x^2 + 2) \div (x - 1) = x^2 - 2x - 2$ Check: $(x - 1)(x^2 - 2x - 2) = x^3 - 2x^2 - 2x - x^2 + 2x + 2 = x^3 - 3x^2 + 2$ ✓ Quadratic $x^2 - 2x - 2$ does not factorise further over integers (discriminant $= 4 + 8 = 12$, not perfect square).

Answer: $(x - 1)(x^2 - 2x - 2)$ [1]

Marking note: 1 mark each. For (b), must show division or inspection method. Accept $(x - 1)(x^2 - 2x - 2)$ as complete factorisation over integers/rationals.

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Section C (Questions 16–20, 3 marks each)

16. $f(x) = \frac{2x + 1}{x - 3}$, $x \neq 3$

(a) Find $f^{-1}(x)$ and its domain.

Working: Let $y = \frac{2x + 1}{x - 3}$ Swap: $x = \frac{2y + 1}{y - 3}$ $x(y - 3) = 2y + 1$ $xy - 3x = 2y + 1$ $xy - 2y = 3x + 1$ $y(x - 2) = 3x + 1$ $y = \frac{3x + 1}{x - 2}$

Domain of $f^{-1}$ = Range of $f$. $f(x) = 2 + \frac{7}{x - 3}$, so $f(x) \neq 2$. Domain: $x \neq 2$.

Answer: $f^{-1}(x) = \frac{3x + 1}{x - 2}$, Domain: $x \neq 2$ [2]

(b) Solve $f(x) = f^{-1}(x)$.

Working: For a function and its inverse, $f(x) = f^{-1}(x)$ implies $f(x) = x$ (intersection on line $y = x$). $\frac{2x + 1}{x - 3} = x$ $2x + 1 = x(x - 3) = x^2 - 3x$ $x^2 - 5x - 1 = 0$ $x = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2}$

Check: Neither solution is 3 (excluded from domain of $f$) or 2 (excluded from domain of $f^{-1}$). Both valid.

Answer: $x = \frac{5 + \sqrt{29}}{2}$ or $x = \frac{5 - \sqrt{29}}{2}$ [1]

Marking note: (a) 2 marks: 1 for correct inverse expression, 1 for correct domain. (b) 1 mark for both solutions. Must solve $f(x) = x$ not $f(x) = f^{-1}(x)$ directly (though they are equivalent here). Common trap: forgetting to check excluded values.

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17. $f(x) = x^2 - 6x + 10$ for $x \ge 3$

(a) Express as $(x - a)^2 + b$.

Working: $f(x) = (x - 3)^2 + 1$

Answer: $(x - 3)^2 + 1$ [1]

(b) Find $f^{-1}(x)$ and its domain.

Working: $y = (x - 3)^2 + 1$, $x \ge 3$ Swap: $x = (y - 3)^2 + 1$ $(y - 3)^2 = x - 1$ $y - 3 = \sqrt{x - 1}$ (positive root since $y \ge 3$) $y = 3 + \sqrt{x - 1}$

Domain of $f^{-1}$ = Range of $f$. Minimum of $f$ is 1 at $x = 3$, so range is $[1, \infty)$. Domain: $x \ge 1$.

Answer: $f^{-1}(x) = 3 + \sqrt{x - 1}$, Domain: $x \ge 1$ [2]

Marking note: (a) 1 mark. (b) 2 marks: 1 for correct inverse with positive root, 1 for correct domain. Must justify positive root choice.

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18. $f(x) = a(x - 2)^2 + 3$, passes through $(4, 11)$

(a) Find $a$.

Working: $11 = a(4 - 2)^2 + 3 = 4a + 3 \Rightarrow 4a = 8 \Rightarrow a = 2$

Answer: $2$ [1]

(b) $g(x) = f(x) - 6$. Describe transformation mapping $y = f(x)$ to $y = g(x)$.

Explanation: Subtracting 6 from the function translates the graph vertically downwards by 6 units.

Answer: Translation of 6 units in the negative $y$-direction (or downwards by 6 units) [1]

(c) Range of $g$ for $x \ge 2$.

Working: $f(x) = 2(x - 2)^2 + 3$, vertex $(2, 3)$, minimum 3 for $x \ge 2$. $g(x) = f(x) - 6 = 2(x - 2)^2 - 3$, minimum $-3$ at $x = 2$. Range: $g(x) \ge -3$ or $[-3, \infty)$

Answer: $g(x) \ge -3$ or $[-3, \infty)$ [1]

Marking note: (a) 1 mark. (b) 1 mark: must use "translation" and specify direction and magnitude. (c) 1 mark: follow-through from (a) and (b).

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19. $f(x) = 2x + 5$, $g(x) = \frac{x - 5}{2}$

(a) Prove $f$ and $g$ are inverses.

Working: $fg(x) = f(g(x)) = f\left(\frac{x - 5}{2}\right) = 2\left(\frac{x - 5}{2}\right) + 5 = x - 5 + 5 = x$ $gf(x) = g(f(x)) = g(2x + 5) = \frac{(2x + 5) - 5}{2} = \frac{2x}{2} = x$ Since $fg(x) = x$ and $gf(x) = x$ for all real $x$, $f$ and $g$ are inverse functions.

Answer: Shown [2]

(b) $h(x) = f(x^2)$. Find $h(-3)$.

Working: $h(-3) = f((-3)^2) = f(9) = 2(9) + 5 = 23$

Answer: $23$ [1]

Marking note: (a) 2 marks: must show both $fg(x) = x$ and $gf(x) = x$. 1 mark each composition 1 mark. (b) 1 mark: careful with $(-3)^2 = 9$, not $-9$.

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20. $f(x) = -x^2 + 4x + 5$

(a) Express as $-(x - a)^2 + b$.

Working: $f(x) = -(x^2 - 4x) + 5$ $= -[(x - 2)^2 - 4] + 5$ $= -(x - 2)^2 + 4 + 5$ $= -(x - 2)^2 + 9$

Answer: $-(x - 2)^2 + 9$ [1]

(b) $f(x) = k$ has exactly one real solution. Find $k$.

Explanation: $f(x) = k \Rightarrow -(x - 2)^2 + 9 = k \Rightarrow (x - 2)^2 = 9 - k$. Exactly one real solution when RHS = 0 $\Rightarrow 9 - k = 0 \Rightarrow k = 9$. (Alternatively: $k$ equals the maximum value of $f$, which is 9 at vertex.)

Answer: $9$ [1]

(c) $g(x) = f(x) + 3 = -(x - 2)^2 + 12$ for $x \le 2$. Explain why $g$ has an inverse, and find $g^{-1}(x)$.

Explanation: For $x \le 2$, $g(x)$ is strictly increasing (left side of vertex, parabola opens downwards). Strictly monotonic $\Rightarrow$ one-to-one $\Rightarrow$ inverse exists.

Working for inverse: $y = -(x - 2)^2 + 12$, $x \le 2$ Swap: $x = -(y - 2)^2 + 12$ $(y - 2)^2 = 12 - x$ $y - 2 = -\sqrt{12 - x}$ (negative root since $y \le 2$) $y = 2 - \sqrt{12 - x}$

Answer: $g$ is strictly increasing on $x \le 2$ (one-to-one). $g^{-1}(x) = 2 - \sqrt{12 - x}$ [1]

Marking note: (a) 1 mark. (b) 1 mark. (c) 1 mark: must explain why inverse exists (strictly monotonic on restricted domain) AND give correct inverse with negative root. Domain of $g^{-1}$ is $x \le 12$ (range of $g$) but not explicitly asked.

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End of Answer Key