Secondary 4 Elementary Mathematics Algebra Functions Quiz

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Secondary 4 Elementary Mathematics Quiz - Algebra Functions

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ________ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working clearly.
• Omission of essential working will result in loss of marks.
• The use of an approved scientific calculator is expected, where appropriate.
• If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.
• For $\pi$, use either your calculator value or 3.142, unless the question requires the answer in terms of $\pi$.

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Section A (10 marks)

Answer all questions. Each question carries 1 mark.

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

Answer: ___________________________ [1]

2. The function $g$ is defined by $g(x) = \frac{2x+1}{x-3}$ for $x \neq 3$. Find $g(5)$.

Answer: ___________________________ [1]

3. If $h(x) = \sqrt{x+4}$, state the domain of $h$.

Answer: ___________________________ [1]

4. The graph of $y = f(x)$ passes through the point $(2, 7)$. The graph of $y = f(x) + 3$ passes through the point $(2, k)$. Find the value of $k$.

Answer: ___________________________ [1]

5. Given that $f(x) = 2x - 1$ and $g(x) = x^2 + 3$, find $fg(2)$.

Answer: ___________________________ [1]

6. The function $f$ is defined by $f(x) = 5 - 2x$ for all real $x$. Find the value of $x$ for which $f(x) = 9$.

Answer: ___________________________ [1]

7. A function $f$ is defined by $f(x) = x^2 - 6x + 10$ for $x \geq 3$. Explain why the inverse function $f^{-1}$ exists.

Answer: ___________________________ [1]

8. The diagram shows the graph of $y = f(x)$ for $-2 \leq x \leq 4$.

Image pending generation: graph for Q8.

Write down the range of $f$ for this domain.

Answer: ___________________________ [1]

9. The function $f$ is defined by $f(x) = \frac{1}{x-2}$ for $x \neq 2$. Write down the equation of the vertical asymptote of the graph $y = f(x)$.

Answer: ___________________________ [1]

10. Given that $f(x) = 3x + 2$ and $g(x) = x^2 - 1$, find $gf(x)$ in terms of $x$, simplifying your answer.

Answer: ___________________________ [1]

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Section B (18 marks)

Answer all questions.

11. The function $f$ is defined by $f(x) = 2x^2 - 8x + 5$ for all real $x$.

(a) Express $f(x)$ in the form $a(x-h)^2 + k$.
Answer: ___________________________ [2]

(b) Hence, write down the coordinates of the vertex of the graph $y = f(x)$.
Answer: ___________________________ [1]

(c) State the minimum value of $f(x)$.
Answer: ___________________________ [1]

12. The function $g$ is defined by $g(x) = \frac{3x-2}{x+1}$ for $x \neq -1$.

(a) Find $g^{-1}(x)$, the inverse function of $g$.
Answer: ___________________________ [3]

(b) State the value of $x$ which must be excluded from the domain of $g^{-1}$.
Answer: ___________________________ [1]

13. Functions $f$ and $g$ are defined by $f(x) = 4x - 3$ and $g(x) = 2x^2 + 1$ for all real $x$.

(a) Find $fg(x)$ in terms of $x$, simplifying your answer.
Answer: ___________________________ [2]

(b) Solve the equation $fg(x) = 29$.
Answer: ___________________________ [2]

14. The function $h$ is defined by $h(x) = \frac{5}{x-2} + 3$ for $x \neq 2$.

(a) Write down the equations of the vertical and horizontal asymptotes of the graph $y = h(x)$.
Answer: Vertical: _______________ Horizontal: _______________ [2]

(b) Sketch the graph of $y = h(x)$ on the axes below, indicating clearly the asymptotes and the intercepts with the axes.

Image pending generation: graph for Q14.

[3]

15. A function $f$ is defined by $f(x) = x^2 - 4x + 7$ for $x \geq 2$.

(a) Show that $f(x) = (x-2)^2 + 3$.
Answer: ___________________________ [1]

(b) Find an expression for $f^{-1}(x)$ and state its domain.
Answer: $f^{-1}(x) =$ ___________________________ Domain: ___________________________ [3]

(c) On the same axes, sketch the graphs of $y = f(x)$ and $y = f^{-1}(x)$, indicating the line $y = x$.

Image pending generation: graph for Q15.

[3]

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Section C (12 marks)

Answer all questions.

16. The function $f$ is defined by $f(x) = ax^2 + bx + c$ for all real $x$, where $a$, $b$, and $c$ are constants. The graph of $y = f(x)$ has a minimum value of $-4$ when $x = 3$, and passes through the point $(0, 5)$.

(a) Write down the value of $c$.
Answer: ___________________________ [1]

(b) Express $f(x)$ in the form $a(x-3)^2 - 4$.
Answer: ___________________________ [1]

(c) Find the values of $a$ and $b$.
Answer: $a =$ _______________ $b =$ _______________ [2]

(d) Find the range of $f$.
Answer: ___________________________ [1]

17. The function $g$ is defined by $g(x) = \frac{2x+5}{x-1}$ for $x \neq 1$.

(a) Find $g^{-1}(x)$.
Answer: ___________________________ [3]

(b) The function $h$ is defined by $h(x) = x^2 - 4$ for $x \geq 0$. Find $gh(3)$.
Answer: ___________________________ [2]

(c) Solve the equation $g(x) = h(x)$ for $x \geq 0$.
Answer: ___________________________ [3]

18. The diagram shows the graph of $y = f(x)$ where $f(x) = \frac{k}{x} + 2$ for $x \neq 0$, and $k$ is a positive constant. The graph passes through the point $P(2, 5)$.

Image pending generation: graph for Q18.

(a) Find the value of $k$.
Answer: ___________________________ [1]

(b) Write down the equations of the asymptotes of the graph.
Answer: ___________________________ [1]

(c) The line $y = 4$ intersects the graph at point $Q$. Find the coordinates of $Q$.
Answer: ___________________________ [2]

(d) On the same axes, sketch the graph of $y = f^{-1}(x)$, indicating clearly the asymptotes and the image of point $P$.
Answer: ___________________________ [3]

19. A function $f$ is defined by $f(x) = 2x^2 - 12x + 19$ for $x \in \mathbb{R}$.

(a) Express $f(x)$ in the form $a(x-h)^2 + k$.
Answer: ___________________________ [2]

(b) The function $g$ is defined by $g(x) = f(x+3)$ for $x \in \mathbb{R}$. Write down the coordinates of the vertex of the graph $y = g(x)$.
Answer: ___________________________ [2]

(c) The function $h$ is defined by $h(x) = f(x) + 5$ for $x \in \mathbb{R}$. Describe fully the single transformation that maps the graph of $y = f(x)$ onto the graph of $y = h(x)$.
Answer: ___________________________ [2]

20. The functions $f$ and $g$ are defined by $f(x) = \sqrt{x-1}$ for $x \geq 1$ and $g(x) = x^2 + 2$ for $x \in \mathbb{R}$.

(a) Find $fg(x)$ in terms of $x$, stating its domain.
Answer: $fg(x) =$ ___________________________ Domain: ___________________________ [2]

(b) Find $gf(x)$ in terms of $x$, stating its domain.
Answer: $gf(x) =$ ___________________________ Domain: ___________________________ [2]

(c) Explain why $fg(x) \neq gf(x)$ in general.
Answer: ___________________________ [1]

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End of Quiz

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Secondary 4 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

Total Marks: 40

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Section A (10 marks)

1. $f(-2) = 3(-2)^2 - 4(-2) + 5 = 3(4) + 8 + 5 = 12 + 8 + 5 = 25$
Answer: 25 [1]

2. $g(5) = \frac{2(5)+1}{5-3} = \frac{10+1}{2} = \frac{11}{2} = 5.5$
Answer: $\frac{11}{2}$ or 5.5 [1]

3. For $h(x) = \sqrt{x+4}$, the expression under the square root must be non-negative: $x+4 \geq 0 \Rightarrow x \geq -4$.
Answer: $x \geq -4$ or $[-4, \infty)$ [1]

4. $y = f(x) + 3$ is a vertical translation of $y = f(x)$ upwards by 3 units. The point $(2, 7)$ maps to $(2, 7+3) = (2, 10)$. So $k = 10$.
Answer: 10 [1]

5. $fg(2) = f(g(2))$. First $g(2) = 2^2 + 3 = 7$. Then $f(7) = 2(7) - 1 = 14 - 1 = 13$.
Answer: 13 [1]

6. $5 - 2x = 9 \Rightarrow -2x = 4 \Rightarrow x = -2$.
Answer: -2 [1]

7. For $x \geq 3$, $f(x) = x^2 - 6x + 10 = (x-3)^2 + 1$. This is a quadratic with vertex at $(3, 1)$ opening upwards. On the restricted domain $x \geq 3$, the function is strictly increasing (one-to-one), so the inverse exists.
Answer: $f$ is one-to-one on $x \geq 3$ (strictly increasing) [1]

8. From the graph, the vertex is at $(1, -3)$ (minimum) and the endpoints are $(-2, 6)$ and $(4, 6)$ (maximum). Range is $[-3, 6]$.
Answer: $-3 \leq y \leq 6$ or $[-3, 6]$ [1]

9. Vertical asymptote occurs where denominator is zero: $x - 2 = 0 \Rightarrow x = 2$.
Answer: $x = 2$ [1]

10. $gf(x) = g(f(x)) = g(3x+2) = (3x+2)^2 - 1 = 9x^2 + 12x + 4 - 1 = 9x^2 + 12x + 3$.
Answer: $9x^2 + 12x + 3$ [1]

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Section B (18 marks)

11. (a) $f(x) = 2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2[(x-2)^2 - 4] + 5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3$
Answer: $2(x-2)^2 - 3$ [2]
Marking: 1 mark for factorising 2, 1 mark for completing square correctly

(b) Vertex is at $(h, k) = (2, -3)$.
Answer: $(2, -3)$ [1]

(c) Minimum value is $k = -3$ (since $a=2>0$, parabola opens upwards).
Answer: -3 [1]

12. (a) Let $y = \frac{3x-2}{x+1}$. Swap $x$ and $y$: $x = \frac{3y-2}{y+1}$.
$x(y+1) = 3y-2 \Rightarrow xy + x = 3y - 2 \Rightarrow xy - 3y = -x - 2 \Rightarrow y(x-3) = -(x+2)$
$y = \frac{-(x+2)}{x-3} = \frac{x+2}{3-x}$
Answer: $g^{-1}(x) = \frac{x+2}{3-x}$ [3]
Marking: 1 mark for swapping, 1 mark for algebraic manipulation, 1 mark for correct final expression

(b) Domain of $g^{-1}$ excludes $x = 3$ (denominator zero). This corresponds to the horizontal asymptote of $g$.
Answer: $x = 3$ [1]

13. (a) $fg(x) = f(g(x)) = f(2x^2+1) = 4(2x^2+1) - 3 = 8x^2 + 4 - 3 = 8x^2 + 1$.
Answer: $8x^2 + 1$ [2]
Marking: 1 mark for substitution, 1 mark for simplification

(b) $8x^2 + 1 = 29 \Rightarrow 8x^2 = 28 \Rightarrow x^2 = \frac{28}{8} = \frac{7}{2} \Rightarrow x = \pm\sqrt{\frac{7}{2}} = \pm\frac{\sqrt{14}}{2}$.
Answer: $x = \frac{\sqrt{14}}{2}$ or $x = -\frac{\sqrt{14}}{2}$ [2]
Marking: 1 mark for setting up equation, 1 mark for solving correctly

14. (a) Vertical asymptote: $x - 2 = 0 \Rightarrow x = 2$.
Horizontal asymptote: As $x \to \pm\infty$, $\frac{5}{x-2} \to 0$, so $y \to 3$.
Answer: Vertical: $x = 2$; Horizontal: $y = 3$ [2]

(b) Graph sketch description:

• Vertical asymptote $x = 2$ (dashed line)
• Horizontal asymptote $y = 3$ (dashed line)
• $y$-intercept: $x=0 \Rightarrow y = \frac{5}{-2} + 3 = -2.5 + 3 = 0.5$ → $(0, 0.5)$
• $x$-intercept: $y=0 \Rightarrow \frac{5}{x-2} = -3 \Rightarrow 5 = -3(x-2) \Rightarrow 5 = -3x+6 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}$ → $(\frac{1}{3}, 0)$
• For $x > 2$, $y > 3$ (upper right branch)
• For $x < 2$, $y < 3$ (lower left branch)
  Answer: See sketch description above [3]
  Marking: 1 mark for asymptotes, 1 mark for intercepts, 1 mark for correct shape and position of branches

15. (a) $x^2 - 4x + 7 = (x^2 - 4x + 4) + 3 = (x-2)^2 + 3$. ✓
Answer: Shown [1]

(b) Let $y = (x-2)^2 + 3$ for $x \geq 2$. Swap: $x = (y-2)^2 + 3$ for $y \geq 2$.
$(y-2)^2 = x - 3 \Rightarrow y-2 = \sqrt{x-3}$ (positive root since $y \geq 2$)
$y = 2 + \sqrt{x-3}$
Domain of $f^{-1}$ = Range of $f$ = $[3, \infty)$ (since minimum is 3 at $x=2$).
Answer: $f^{-1}(x) = 2 + \sqrt{x-3}$, Domain: $x \geq 3$ [3]
Marking: 1 mark for correct algebraic manipulation, 1 mark for choosing correct root, 1 mark for domain

(c) Graph sketch description:

• $y = f(x) = (x-

<stage5_quiz_answers_md>

Secondary 4 Elementary Mathematics Quiz - Algebra Functions (Answer Key)

Total Marks: 40

---

Section A (10 marks)

1. $f(-2) = 3(-2)^2 - 4(-2) + 5 = 3(4) + 8 + 5 = 12 + 8 + 5 = 25$
Answer: 25 [1]

2. $g(5) = \frac{2(5)+1}{5-3} = \frac{10+1}{2} = \frac{11}{2} = 5.5$
Answer: $\frac{11}{2}$ or 5.5 [1]

3. For $h(x) = \sqrt{x+4}$, the expression under the square root must be non-negative: $x+4 \geq 0 \Rightarrow x \geq -4$.
Answer: $x \geq -4$ or $[-4, \infty)$ [1]

4. $y = f(x) + 3$ is a vertical translation of $y = f(x)$ upwards by 3 units. The point $(2, 7)$ maps to $(2, 7+3) = (2, 10)$. So $k = 10$.
Answer: 10 [1]

5. $fg(2) = f(g(2))$. First $g(2) = 2^2 + 3 = 7$. Then $f(7) = 2(7) - 1 = 14 - 1 = 13$.
Answer: 13 [1]

6. $5 - 2x = 9 \Rightarrow -2x = 4 \Rightarrow x = -2$.
Answer: -2 [1]

7. For $x \geq 3$, $f(x) = x^2 - 6x + 10 = (x-3)^2 + 1$. This is a quadratic with vertex at $(3, 1)$ opening upwards. On the restricted domain $x \geq 3$, the function is strictly increasing (one-to-one), so the inverse exists.
Answer: $f$ is one-to-one on $x \geq 3$ (strictly increasing) [1]

8. From the graph, the vertex is at $(1, -3)$ (minimum) and the endpoints are $(-2, 6)$ and $(4, 6)$ (maximum). Range is $[-3, 6]$.
Answer: $-3 \leq y \leq 6$ or $[-3, 6]$ [1]

9. Vertical asymptote occurs where denominator is zero: $x - 2 = 0 \Rightarrow x = 2$.
Answer: $x = 2$ [1]

10. $gf(x) = g(f(x)) = g(3x+2) = (3x+2)^2 - 1 = 9x^2 + 12x + 4 - 1 = 9x^2 + 12x + 3$.
Answer: $9x^2 + 12x + 3$ [1]

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Section B (18 marks)

11. (a) $f(x) = 2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2[(x-2)^2 - 4] + 5 = 2(x-2)^2 - 8 + 5 = 2(x-2)^2 - 3$
Answer: $2(x-2)^2 - 3$ [2]
Marking: 1 mark for factorising 2, 1 mark for completing square correctly

(b) Vertex is at $(h, k) = (2, -3)$.
Answer: $(2, -3)$ [1]

(c) Minimum value is $k = -3$ (since $a=2>0$, parabola opens upwards).
Answer: -3 [1]

12. (a) Let $y = \frac{3x-2}{x+1}$. Swap $x$ and $y$: $x = \frac{3y-2}{y+1}$.
$x(y+1) = 3y-2 \Rightarrow xy + x = 3y - 2 \Rightarrow xy - 3y = -x - 2 \Rightarrow y(x-3) = -(x+2)$
$y = \frac{-(x+2)}{x-3} = \frac{x+2}{3-x}$
Answer: $g^{-1}(x) = \frac{x+2}{3-x}$ [3]
Marking: 1 mark for swapping, 1 mark for algebraic manipulation, 1 mark for correct final expression

(b) Domain of $g^{-1}$ excludes $x = 3$ (denominator zero). This corresponds to the horizontal asymptote of $g$.
Answer: $x = 3$ [1]

13. (a) $fg(x) = f(g(x)) = f(2x^2+1) = 4(2x^2+1) - 3 = 8x^2 + 4 - 3 = 8x^2 + 1$.
Answer: $8x^2 + 1$ [2]
Marking: 1 mark for substitution, 1 mark for simplification

(b) $8x^2 + 1 = 29 \Rightarrow 8x^2 = 28 \Rightarrow x^2 = \frac{28}{8} = \frac{7}{2} \Rightarrow x = \pm\sqrt{\frac{7}{2}} = \pm\frac{\sqrt{14}}{2}$.
Answer: $x = \frac{\sqrt{14}}{2}$ or $x = -\frac{\sqrt{14}}{2}$ [2]
Marking: 1 mark for setting up equation, 1 mark for solving correctly

14. (a) Vertical asymptote: $x - 2 = 0 \Rightarrow x = 2$.
Horizontal asymptote: As $x \to \pm\infty$, $\frac{5}{x-2} \to 0$, so $y \to 3$.
Answer: Vertical: $x = 2$; Horizontal: $y = 3$ [2]

(b) Graph sketch description:

• Vertical asymptote $x = 2$ (dashed line)
• Horizontal asymptote $y = 3$ (dashed line)
• $y$-intercept: $x=0 \Rightarrow y = \frac{5}{-2} + 3 = -2.5 + 3 = 0.5$ → $(0, 0.5)$
• $x$-intercept: $y=0 \Rightarrow \frac{5}{x-2} = -3 \Rightarrow 5 = -3(x-2) \Rightarrow 5 = -3x+6 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}$ → $(\frac{1}{3}, 0)$
• For $x > 2$, $y > 3$ (upper right branch)
• For $x < 2$, $y < 3$ (lower left branch)
  Answer: See sketch description above [3]
  Marking: 1 mark for asymptotes, 1 mark for intercepts, 1 mark for correct shape and position of branches

15. (a) $x^2 - 4x + 7 = (x^2 - 4x + 4) + 3 = (x-2)^2 + 3$. ✓
Answer: Shown [1]

(b) Let $y = (x-2)^2 + 3$ for $x \geq 2$. Swap: $x = (y-2)^2 + 3$ for $y \geq 2$.
$(y-2)^2 = x - 3 \Rightarrow y-2 = \sqrt{x-3}$ (positive root since $y \geq 2$)
$y = 2 + \sqrt{x-3}$
Domain of $f^{-1}$ = Range of $f$ = $[3, \infty)$ (since minimum is 3 at $x=2$).
Answer: $f^{-1}(x) = 2 + \sqrt{x-3}$, Domain: $x \geq 3$ [3]
Marking: 1 mark for correct algebraic manipulation, 1 mark for choosing correct root, 1 mark for domain

(c) Graph sketch description:

• $y = f(x) = (x-2)^2 + 3$ for $x \geq 2$: parabola vertex at $(2, 3)$, opening upwards, starting at $(2, 3)$
• $y = f^{-1}(x) = 2 + \sqrt{x-3}$ for $x \geq 3$: square root curve starting at $(3, 2)$, increasing
• Line $y = x$ (dashed)
• Intersection points: $(3, 3)$ and $(4, 4)$
• Graphs are reflections of each other across $y = x$
  Answer: See sketch description above [3]
  Marking: 1 mark for $f(x)$ correct shape and domain, 1 mark for $f^{-1}(x)$ correct shape and domain, 1 mark for line $y=x$ and reflection symmetry

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Section C (12 marks)

16. (a) $f(0) = c = 5$ (since graph passes through $(0, 5)$).
Answer: $c = 5$ [1]

(b) Minimum value $-4$ at $x=3$ → vertex form: $f(x) = a(x-3)^2 - 4$.
Answer: $a(x-3)^2 - 4$ [1]

(c) Using $f(0) = 5$: $a(0-3)^2 - 4 = 5 \Rightarrow 9a - 4 = 5 \Rightarrow 9a = 9 \Rightarrow a = 1$.
Then $f(x) = (x-3)^2 - 4 = x^2 - 6x + 9 - 4 = x^2 - 6x + 5$. So $b = -6$.
Answer: $a = 1$, $b = -6$ [2]
Marking: 1 mark for finding $a$, 1 mark for finding $b$

(d) Since $a=1>0$, minimum is $-4$. Range: $[-4, \infty)$ or $f(x) \geq -4$.
Answer: $f(x) \geq -4$ or $[-4, \infty)$ [1]

17. (a) Let $y = \frac{2x+5}{x-1}$. Swap: $x = \frac{2y+5}{y-1}$.
$x(y-1) = 2y+5 \Rightarrow xy - x = 2y + 5 \Rightarrow xy - 2y = x + 5 \Rightarrow y(x-2) = x+5$
$y = \frac{x+5}{x-2}$
Answer: $g^{-1}(x) = \frac{x+5}{x-2}$ [3]
Marking: 1 mark for swapping, 1 mark for algebra, 1 mark for final expression

(b) $h(3) = 3^2 - 4 = 9 - 4 = 5$.
$g(5) = \frac{2(5)+5}{5-1} = \frac{10+5}{4} = \frac{15}{4} = 3.75$.
Answer: $\frac{15}{4}$ or 3.75 [2]
Marking: 1 mark for $h(3)$, 1 mark for $g(5)$

(c) $\frac{2x+5}{x-1} = x^2 - 4$ for $x \geq 0$, $x \neq 1$.
$2x+5 = (x^2-4)(x-1) = x^3 - x^2 - 4x + 4$
$x^3 - x^2 - 6x - 1 = 0$
By trial, $x = -1$ is a root but $x \geq 0$. Use numerical methods or factor:
$(x+1)(x^2 - 2x - 1) = 0$ → $x = -1$ or $x = 1 \pm \sqrt{2}$.
For $x \geq 0$: $x = 1 + \sqrt{2}$ (since $1 - \sqrt{2} < 0$).
Answer: $x = 1 + \sqrt{2}$ [3]
Marking: 1 mark for setting up equation, 1 mark for simplifying to cubic, 1 mark for correct root in domain

18. (a) $f(2) = \frac{k}{2} + 2 = 5 \Rightarrow \frac{k}{2} = 3 \Rightarrow k = 6$.
Answer: $k = 6$ [1]

(b) Vertical asymptote: $x = 0$. Horizontal asymptote: $y = 2$.
Answer: $x = 0$ and $y = 2$ [1]

(c) $y = 4$: $\frac{6}{x} + 2 = 4 \Rightarrow \frac{6}{x} = 2 \Rightarrow x = 3$.
Coordinates of $Q$: $(3, 4)$.
Answer: $(3, 4)$ [2]
Marking: 1 mark for equation, 1 mark for coordinates

(d) $f(x) = \frac{6}{x} + 2$. Find inverse: $y = \frac{6}{x} + 2 \Rightarrow x = \frac{6}{y} + 2 \Rightarrow x-2 = \frac{6}{y} \Rightarrow y = \frac{6}{x-2}$.
$f^{-1}(x) = \frac{6}{x-2}$.
Asymptotes of $f^{-1}$: vertical $x = 2$, horizontal $y = 0$.
Image of $P(2, 5)$ under reflection in $y=x$ is $P'(5, 2)$.
Answer: $f^{-1}(x) = \frac{6}{x-2}$; asymptotes $x=2$, $y=0$; $P'(5, 2)$ [3]
Marking: 1 mark for inverse function, 1 mark for asymptotes, 1 mark for image of P

19. (a) $f(x) = 2x^2 - 12x + 19 = 2(x^2 - 6x) + 19 = 2[(x-3)^2 - 9] + 19 = 2(x-3)^2 - 18 + 19 = 2(x-3)^2 + 1$.
Answer: $2(x-3)^2 + 1$ [2]
Marking: 1 mark for factorising 2, 1 mark for completing square

(b) $g(x) = f(x+3) = 2((x+3)-3)^2 + 1 = 2x^2 + 1$. Vertex at $(0, 1)$.
Answer: $(0, 1)$ [2]
Marking: 1 mark for expression, 1 mark for vertex

(c) $h(x) = f(x) + 5$. This is a translation of $y = \begin{pmatrix} 0 \\ 5 \end{pmatrix}$ (5 units upwards).
Answer: Translation by vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ (5 units up) [2]
Marking: 1 mark for "translation", 1 mark for correct vector/direction

20. (a) $fg(x) = f(g(x)) = f(x^2+2) = \sqrt{(x^2+2)-1} = \sqrt{x^2+1}$.
Domain of $g$: $\mathbb{R}$. Range of $g$: $[2, \infty)$. Domain of $f$: $[1, \infty)$. Since range of $g \subseteq$ domain of $f$, domain of $fg$ is $\mathbb{R}$.
Answer: $fg(x) = \sqrt{x^2+1}$, Domain: $x \in \mathbb{R}$ [2]
Marking: 1 mark for expression, 1 mark for domain

(b) $gf(x) = g(f(x)) = g(\sqrt{x-1}) = (\sqrt{x-1})^2 + 2 = x-1+2 = x+1$.
Domain of $f$: $x \geq 1$. Range of $f$: $[0, \infty)$. Domain of $g$: $\mathbb{R}$. Since range of $f \subseteq$ domain of $g$, domain of $gf$ is $x \geq 1$.
Answer: $gf(x) = x+1$, Domain: $x \geq 1$ [2]
Marking: 1 mark for expression, 1 mark for domain

(c) $fg(x) = \sqrt{x^2+1}$ while $gf(x) = x+1$. These are different functions (e.g., at $x=0$, $fg(0)=1$, $gf(0)$ undefined; at $x=1$, $fg(1)=\sqrt{2}$, $gf(1)=2$). Function composition is not commutative in general.
Answer: $fg(x) = \sqrt{x^2+1}$ and $gf(x) = x+1$ are different functions; composition is not commutative. [1]

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