This checkpoint assesses your understanding of linear versus non-linear graphs, the basic parabola $y = x^2$, dilations $y = ax^2$, reflections $y = -ax^2$, vertical translations $y = x^2 + c$, the combined form $y = ax^2 + c$, and horizontal translations $y = (x - h)^2$. It covers Lessons 1–7.
❓ Multiple Choice (10 questions)
1. Which of these equations produces a curve (a non-linear graph)?
2. For the basic parabola $y = x^2$, what is the value of $y$ when $x = -4$?
3. What are the coordinates of the vertex of $y = x^2$?
4. Compared with $y = x^2$, the parabola $y = 3x^2$ is:
5. The graph of $y = -2x^2$ opens:
6. What is the vertex of $y = x^2 + 5$?
7. The parabola $y = x^2 - 3$ is the graph of $y = x^2$ moved:
8. What is the vertex of $y = 2x^2 - 6$?
9. What is the vertex of $y = (x - 4)^2$?
10. The equation of a parabola that is the same shape as $y = x^2$ but with vertex at $(-3, 0)$ is:
✍ Short Answer (4 questions)
11. Work with the basic parabola $y = x^2$.
(a) Find $y$ when $x = -2$. (1 mark)
(b) Find $y$ when $x = 3$. (1 mark)
(c) State the coordinates of the vertex and the equation of the axis of symmetry. (1 mark)3 MARKS
12. Describe each parabola compared with $y = x^2$ (direction of opening and whether it is narrower or wider).
(a) $y = 4x^2$ (1 mark)
(b) $y = -x^2$ (1 mark)
(c) $y = \tfrac{1}{3}x^2$ (2 marks)4 MARKS
13. For each parabola, state the coordinates of the vertex and the equation of the axis of symmetry.
(a) $y = x^2 + 7$ (1 mark)
(b) $y = 3x^2 - 2$ (2 marks)
(c) $y = (x - 5)^2$ (2 marks)5 MARKS
14. Write the equation of each parabola described below.
(a) Same shape as $y = x^2$ but moved up 6 units. (1 mark)
(b) Same shape as $y = x^2$ but with vertex at $(2, 0)$. (1 mark)
(c) Opens downward, same width as $y = x^2$, vertex at the origin. (1 mark)
(d) A parabola with vertex $(0, -4)$ that is narrower than $y = x^2$ (give one possible equation). (2 marks)5 MARKS
1. C$y = x^2$ contains a squared variable, so it is a parabola (a curve). The others are all of the form $y = mx + c$ (straight lines).
2. B$y = (-4)^2 = 16$. Squaring removes the negative sign.
3. AThe vertex (lowest point) of $y = x^2$ is at $(0, 0)$.
4. D$a = 3 > 1$, so the parabola is narrower; $a > 0$ so it opens up.
5. B$a = -2 < 0$, so it opens downward and the vertex is a maximum.
6. CAdding $5$ raises the curve. The vertex moves from $(0,0)$ to $(0, 5)$.
7. AThe $-3$ lowers the curve, so it is moved down 3 units (vertex $(0, -3)$).
8. DFor $y = ax^2 + c$ the vertex is $(0, c)$; here $c = -6$, so the vertex is $(0, -6)$.
9. B$(x - 4) = 0$ when $x = 4$, so the vertex is $(4, 0)$ (minus inside means shift right).
10. CA vertex of $(-3, 0)$ needs $(x - (-3))^2 = (x + 3)^2$ (plus inside means shift left).
Q11 (3 marks): (a) $y = (-2)^2 = 4$ [1]. (b) $y = (3)^2 = 9$ [1]. (c) Vertex $(0, 0)$; axis of symmetry $x = 0$ [1].
Q12 (4 marks): (a) $a = 4 > 1$: opens up, narrower than $y = x^2$ [1]. (b) $a = -1$: opens downward, same width [1]. (c) $a = \tfrac{1}{3}$ (between 0 and 1): opens up, wider (flatter) than $y = x^2$ [2].
Q13 (5 marks): (a) Vertex $(0, 7)$, axis $x = 0$ [1]. (b) Vertex $(0, -2)$, axis $x = 0$ [2]. (c) $(x - 5) = 0$ at $x = 5$, so vertex $(5, 0)$, axis $x = 5$ [2].
Q14 (5 marks): (a) $y = x^2 + 6$ [1]. (b) $y = (x - 2)^2$ [1]. (c) $y = -x^2$ [1]. (d) Any equation of the form $y = ax^2 - 4$ with $a > 1$, e.g. $y = 2x^2 - 4$ [2].
Tick when you have finished all questions and checked your answers.