This quiz covers all 20 lessons of Unit 2: parabolas and their transformations, vertex and intercepts, circles, hyperbolas, exponentials, and solving quadratic equations. It includes 15 multiple choice questions, 3 short answer questions and 1 extended response. Show all working for full marks.
❓ Multiple Choice (15 questions)
1. What is the vertex of the basic parabola $y = x^2$?
2. Compared with $y = x^2$, the parabola $y = 3x^2$ is:
3. What is the vertex of $y = x^2 - 4$?
4. What is the vertex of $y = (x + 2)^2$?
5. The parabola $y = -x^2$ opens:
6. What is the vertex of $y = (x - 1)^2 + 5$?
7. What is the radius of the circle $x^2 + y^2 = 64$?
8. For the hyperbola $y = \dfrac{8}{x}$, what is $y$ when $x = 4$?
9. For $y = 2^x$, what is $y$ when $x = 0$?
10. What is the $y$-intercept of $y = (x - 2)^2 + 3$?
11. What are the $x$-intercepts of $y = x^2 - 25$?
12. Solve $x^2 - 7x + 10 = 0$.
13. Which family of graph is $y = \dfrac{6}{x}$?
14. Solve $x^2 + 5x + 6 = 0$.
15. Does the point $(0, -5)$ lie on the circle $x^2 + y^2 = 25$?
✍ Short Answer (3 questions)
16. Parabolas and transformations.
(a) Describe how $y = 2x^2 - 3$ differs from $y = x^2$ (width, direction and vertex). (2 marks)
(b) State the vertex and axis of symmetry of $y = (x - 4)^2 + 1$. (2 marks)
(c) Find the $y$-intercept of $y = (x + 1)^2 - 6$. (2 marks)6 MARKS
17. The non-linear families.
(a) State the centre and radius of $x^2 + y^2 = 81$. (2 marks)
(b) For $y = \dfrac{10}{x}$, find $y$ when $x = 5$ and state the two asymptotes. (2 marks)
(c) State the $y$-intercept and horizontal asymptote of $y = 3^x$. (1 mark)5 MARKS
18. Quadratic equations.
(a) Solve $x^2 - 9 = 0$. (1 mark)
(b) Solve $x^2 - 2x - 8 = 0$ by factoring. (2 marks)
(c) Find the $x$-intercepts of $y = x^2 + 6x + 8$. (2 marks)5 MARKS
✍ Extended Response
19. A ball is thrown upward. Its height $h$ (in metres) after $t$ seconds is modelled by $h = -t^2 + 8t$.
(a) Factorise $-t^2 + 8t$ and find the times when $h = 0$. (2 marks)
(b) How long is the ball in the air before it lands? (1 mark)
(c) The path is a parabola. Use the axis of symmetry to find the time at which the ball reaches its maximum height, and find that maximum height. (3 marks)
(d) Find the two times when the ball is exactly $15$ m high by solving $-t^2 + 8t = 15$. (2 marks)
(e) Explain, using the symmetry of the parabola, why your two answers in (d) are equally spaced either side of the time found in (c). (2 marks)10 MARKS
1. AThe vertex of $y = x^2$ is at $(0, 0)$.
2. C$a = 3 > 1$, so the parabola is narrower, and $a > 0$ so it opens up.
3. BFor $y = x^2 + c$ the vertex is $(0, c)$; here $c = -4$, so $(0, -4)$.
4. C$(x + 2) = 0$ at $x = -2$, so the vertex is $(-2, 0)$ (plus inside means shift left).
5. A$a = -1 < 0$, so it opens downward and the vertex is a maximum.
6. DVertex form $y = a(x - h)^2 + k$ has vertex $(h, k)$; here $h = 1$, $k = 5$, so $(1, 5)$.
7. B$r^2 = 64$, so $r = \sqrt{64} = 8$.
8. C$y = \dfrac{8}{4} = 2$.
9. A$2^0 = 1$ (any non-zero base to the power 0 is 1).
10. DSub $x = 0$: $y = (0 - 2)^2 + 3 = 4 + 3 = 7$, so $(0, 7)$.
11. C$x^2 = 25 \Rightarrow x = \pm 5$, so $x = 5$ and $x = -5$.
12. B$x^2 - 7x + 10 = (x - 2)(x - 5) = 0$, so $x = 2$ or $x = 5$.
13. A$y = \dfrac{k}{x}$ is a hyperbola.
14. D$x^2 + 5x + 6 = (x + 2)(x + 3) = 0$, so $x = -2$ or $x = -3$.
15. C$0^2 + (-5)^2 = 0 + 25 = 25 = r^2$, so the point lies on the circle.
Q16 (6 marks): (a) $a = 2 > 1$: narrower and opens up; $c = -3$ moves it down so the vertex is at $(0, -3)$ [2]. (b) Vertex $(4, 1)$, axis of symmetry $x = 4$ [2]. (c) Sub $x = 0$: $y = (0 + 1)^2 - 6 = 1 - 6 = -5$, so $(0, -5)$ [2].
Q17 (5 marks): (a) Centre $(0, 0)$, radius $r = \sqrt{81} = 9$ [2]. (b) $y = \dfrac{10}{5} = 2$; asymptotes $x = 0$ and $y = 0$ [2]. (c) $y$-intercept $(0, 1)$, asymptote $y = 0$ [1].
Q18 (5 marks): (a) $x^2 = 9 \Rightarrow x = \pm 3$ [1]. (b) $(x - 4)(x + 2) = 0 \Rightarrow x = 4$ or $x = -2$ [2]. (c) $x^2 + 6x + 8 = (x + 2)(x + 4) = 0 \Rightarrow x = -2$ or $x = -4$, so $x$-intercepts $(-2, 0)$ and $(-4, 0)$ [2].
Q19 (10 marks):
(a) $-t^2 + 8t = -t(t - 8)$. Setting $h = 0$: $-t(t - 8) = 0 \Rightarrow t = 0$ or $t = 8$ [2].
(b) The ball is at ground level at $t = 0$ (launch) and $t = 8$ (landing), so it is in the air for 8 seconds [1].
(c) The axis of symmetry lies midway between the roots $t = 0$ and $t = 8$, so $t = \dfrac{0 + 8}{2} = 4$ s. Maximum height: $h = -(4)^2 + 8(4) = -16 + 32 = 16$ m [3].
(d) $-t^2 + 8t = 15 \Rightarrow t^2 - 8t + 15 = 0 \Rightarrow (t - 3)(t - 5) = 0 \Rightarrow t = 3$ s or $t = 5$ s [2].
(e) A parabola is symmetric about its axis of symmetry ($t = 4$). At the same height the ball is at two times that are mirror images across $t = 4$: $3$ is $1$ second before $4$ and $5$ is $1$ second after $4$, so the two times are equally spaced either side of the maximum [2].
Tick when you have finished all questions and checked your answers.