This checkpoint assesses your understanding of the vertex and axis of symmetry, finding $x$- and $y$-intercepts, sketching and identifying parabolas, and reading the general (vertex) form $y = a(x - h)^2 + k$. It covers Lessons 8–11.
❓ Multiple Choice (10 questions)
1. What is the vertex of $y = (x - 2)^2 + 3$?
2. What is the equation of the axis of symmetry of $y = (x - 6)^2 + 1$?
3. What is the $y$-intercept of $y = x^2 - 5$?
4. What are the $x$-intercepts of $y = x^2 - 16$?
5. How many $x$-intercepts does $y = x^2 + 9$ have?
6. The vertex of $y = -(x + 4)^2 + 2$ is a:
7. What is the $y$-intercept of $y = (x - 3)^2$?
8. The parabola $y = a(x - h)^2 + k$ has its vertex at $(-1, -7)$. The values of $h$ and $k$ are:
9. A parabola has vertex $(1, -4)$ and passes through $(0, -3)$. Its equation is:
10. What are the $x$-intercepts of $y = (x - 1)^2 - 9$?
✍ Short Answer (4 questions)
11. For each parabola, state the vertex and the equation of the axis of symmetry.
(a) $y = (x - 3)^2 + 4$ (1 mark)
(b) $y = (x + 2)^2 - 5$ (1 mark)
(c) $y = -2(x - 1)^2 + 6$ (1 mark)3 MARKS
12. Find the intercepts of each parabola.
(a) Find the $y$-intercept of $y = x^2 - 7$. (1 mark)
(b) Find the $x$-intercepts of $y = x^2 - 25$. (1 mark)
(c) Find both the $y$-intercept and the $x$-intercepts of $y = (x - 2)^2$. (2 marks)4 MARKS
13. Consider $y = (x - 4)^2 - 1$.
(a) State the vertex and direction of opening. (1 mark)
(b) Find the $y$-intercept. (1 mark)
(c) Find the $x$-intercepts. (2 marks)4 MARKS
14. Identifying parabolas from information given.
(a) A parabola has vertex $(3, 2)$ and the same shape as $y = x^2$. Write its equation in vertex form. (2 marks)
(b) A parabola has vertex $(-2, 1)$ and passes through the point $(0, 9)$. Find the value of $a$ and write the equation $y = a(x - h)^2 + k$. (3 marks)5 MARKS
1. AIn $y = a(x - h)^2 + k$ the vertex is $(h, k)$. Here $h = 2$, $k = 3$, so the vertex is $(2, 3)$.
2. CThe axis of symmetry is $x = h$. Here $h = 6$, so it is $x = 6$.
3. BSubstitute $x = 0$: $y = 0 - 5 = -5$, giving $(0, -5)$.
4. DSet $y = 0$: $x^2 = 16 \Rightarrow x = \pm 4$, so $x = 4$ and $x = -4$.
5. BSet $y = 0$: $x^2 = -9$ has no real solution, so there are no $x$-intercepts (vertex $(0, 9)$ sits above the axis, opening up).
6. C$a = -1 < 0$ so the vertex is a maximum; $(x + 4) = 0$ at $x = -4$, $k = 2$, so vertex $(-4, 2)$.
7. ASubstitute $x = 0$: $y = (0 - 3)^2 = 9$, giving $(0, 9)$.
8. DVertex $(h, k) = (-1, -7)$, so $h = -1$ and $k = -7$.
9. BVertex $(1, -4)$ gives $y = a(x - 1)^2 - 4$. Sub $(0, -3)$: $-3 = a(1) - 4 \Rightarrow a = 1$, so $y = (x - 1)^2 - 4$.
10. CSet $y = 0$: $(x - 1)^2 = 9 \Rightarrow x - 1 = \pm 3 \Rightarrow x = 4$ or $x = -2$.
Q11 (3 marks): (a) Vertex $(3, 4)$, axis $x = 3$ [1]. (b) Vertex $(-2, -5)$, axis $x = -2$ [1]. (c) Vertex $(1, 6)$, axis $x = 1$ [1].
Q12 (4 marks): (a) Sub $x = 0$: $y = -7$, so $(0, -7)$ [1]. (b) $x^2 = 25 \Rightarrow x = \pm 5$, so $(5, 0)$ and $(-5, 0)$ [1]. (c) $y$-int: sub $x = 0$, $y = 4$, so $(0, 4)$. $x$-int: $(x - 2)^2 = 0 \Rightarrow x = 2$ (one repeated root), so $(2, 0)$ [2].
Q13 (4 marks): (a) Vertex $(4, -1)$, opens upward ($a = 1 > 0$) [1]. (b) Sub $x = 0$: $y = (0 - 4)^2 - 1 = 16 - 1 = 15$, so $(0, 15)$ [1]. (c) Set $y = 0$: $(x - 4)^2 = 1 \Rightarrow x - 4 = \pm 1 \Rightarrow x = 5$ or $x = 3$, so $(5, 0)$ and $(3, 0)$ [2].
Q14 (5 marks): (a) Same shape as $y = x^2$ means $a = 1$; vertex $(3, 2)$ gives $y = (x - 3)^2 + 2$ [2]. (b) $y = a(x + 2)^2 + 1$. Sub $(0, 9)$: $9 = a(2)^2 + 1 = 4a + 1 \Rightarrow 4a = 8 \Rightarrow a = 2$. Equation: $y = 2(x + 2)^2 + 1$ [3].
Tick when you have finished all questions and checked your answers.