Year 9 Introduction to quadratic equations and graphs: free practice test with answers

Sample introduction to quadratic equations and graphs questions with answers

Try each one before you open the answer.

1. Quadratic equations

   Which of these are quadratic equations? Give a reason for each.

   (a) $x^2 - 5x = 0$ (b) $3x + 2 = 0$ (c) $x(x + 4) = 7$ (d) $x^3 = 8$

   Show the answer

   Key idea: A quadratic equation can be written as $ax^2 + bx + c = 0$ with $a \ne 0$: the highest power of $x$ is $2$.

   1. (a) The highest power is $x^2$: quadratic
   2. (b) The highest power is $x^1$: linear, not quadratic
   3. (c) Expanding gives $x^2 + 4x = 7$, so the highest power is $x^2$: quadratic
   4. (d) The highest power is $x^3$: not quadratic

   Answer: (a) yes (b) no (c) yes (d) no

2. Solving ax² + bx = 0 and x² = d

   Solve:

   (a) $x^2 = 49$

   (b) $x^2 = 20$, giving exact answers and answers correct to $2$ decimal places

   Show the answer

   Key idea: If $x^2 = d$ with $d > 0$, then $x = \pm\sqrt{d}$: there are two solutions, one positive and one negative.

   1. (a) $x = \pm\sqrt{49} = \pm 7$
   2. (b) $x = \pm\sqrt{20} = \pm\sqrt{4 \times 5} = \pm 2\sqrt{5}$
   3. (b) $2\sqrt{5} = 4.472\ldots \approx 4.47$

   Answer: (a) $x = \pm 7$ (b) $x = \pm 2\sqrt{5} \approx \pm 4.47$

3. Solving x² + bx + c = 0

   Solve $x^2 - 5x + 6 = 0$.

   Show the answer

   Key idea: Factorise the quadratic into two brackets, then use the null factor law: if $a \times b = 0$, then $a = 0$ or $b = 0$.

   1. Find two integers that multiply to $6$ and add to $-5$: these are $-2$ and $-3$.
   2. Factorise: $(x - 2)(x - 3) = 0$
   3. Apply the null factor law:\

   $$x - 2 = 0 \text{ or } x - 3 = 0$$

   4. Solve each equation:\

   $$x = 2 \text{ or } x = 3$$

   5. Check by substituting back: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ ✓ and $(3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ ✓

   Answer: $x = 2$ or $x = 3$

4. Using quadratic equations to solve problems

   A rectangular garden has a length that is $5$ metres more than its width. The area of the garden is $36$ square metres. Find the dimensions of the garden.

   Show the answer

   Key idea: Write an equation for the area of the rectangle, rearrange to equal zero, then factorise and apply the null factor law. Reject any negative solution since a length must be positive.

   1. Let the width be $x$ metres. Then the length is $(x + 5)$ metres.
   2. Set up the area equation:\

   $$x(x + 5) = 36$$

   3. Expand and rearrange:\

   $$x^2 + 5x - 36 = 0$$

   4. Factorise: find two integers that multiply to $-36$ and add to $5$: these are $9$ and $-4$.\

   $$(x + 9)(x - 4) = 0$$

   5. Apply the null factor law: $x = -9$ or $x = 4$. Since width must be positive, $x = 4$.
   6. Width $= 4$ m, Length $= 4 + 5 = 9$ m. Check: $4 \times 9 = 36$ ✓

   Answer: The garden is $4$ metres wide and $9$ metres long.

5. The parabola

   Consider the parabola $y = x^2 - 4x + 3$.

   (a) Copy and complete the table of values.

   $x$	$0$	$1$	$2$	$3$	$4$
   $y$	$\Box$	$\Box$	$\Box$	$\Box$	$\Box$

   (b) Use the table to state the turning point, the axis of symmetry, the $x$-intercepts and the $y$-intercept.

   Show the answer

   Key idea: The table is symmetric about the axis of symmetry, and the turning point lies on that axis. The $x$-intercepts are where $y = 0$.

   1. (a) $x = 0$: $3$; $x = 1$: $1 - 4 + 3 = 0$; $x = 2$: $4 - 8 + 3 = -1$; $x = 3$: $0$; $x = 4$: $3$
   2. (b) The values are symmetric about $x = 2$, so the axis is $x = 2$ and the turning point is $(2, -1)$
   3. (b) $y = 0$ at $x = 1$ and $x = 3$; at $x = 0$, $y = 3$

   Answer: (a) 3, 0, −1, 0, 3 (b) turning point $(2, -1)$, axis $x = 2$, $x$-intercepts $1$ and $3$, $y$-intercept $3$

6. Sketching y = ax² with dilations and reflections

   The graph of $y = x^2$ is transformed to give $y = -3x^2 + 4$. Describe the transformations applied to the standard parabola, state whether the new parabola opens upward or downward, give its vertex, and find the $y$-intercept.

   Show the answer

   Key idea: In $y = ax^2 + c$, the value of $a$ controls the stretch and reflection, and $c$ shifts the graph vertically.

   1. Compare with the standard form $y = ax^2 + c$: here $a = -3$ and $c = 4$.
   2. Reflection: Because $a = -3 < 0$, the parabola is reflected in the $x$-axis (opens downward).
   3. Vertical stretch: $\lvert a \rvert = 3 > 1$, so the parabola is narrower than $y = x^2$ by a factor of $3$.
   4. Vertical translation: $c = 4$, so the graph is shifted $4$ units upward.
   5. The vertex is at $(0,\ 4)$ and the $y$-intercept is also $(0,\ 4)$.

   Answer: Reflected in the $x$-axis, stretched vertically by factor $3$, shifted $4$ units up; opens downward; vertex and $y$-intercept at $(0, 4)$.

7. Sketching translations of y = x²

   For the parabola $y = (x + 5)^{2} - 2$:

   (a) write the coordinates of the turning point

   (b) find the $y$-intercept

   Show the answer

   Key idea: For $y = (x - h)^{2} + k$ the turning point is $(h, k)$.

   1. $y = (x - (-5))^{2} + (-2)$, so the turning point is $(-5, -2)$
   2. When $x = 0$: $y = 5^{2} - 2 = 23$

   Answer: (a) $(-5, -2)$ (b) $23$

8. Sketching parabolas using intercept form

   For the parabola $y = (x - 1)(x - 5)$, find:

   (a) the $x$-intercepts (b) the $y$-intercept (c) the axis of symmetry (d) the turning point

   Show the answer

   Key idea: In intercept form $y = (x - p)(x - q)$, the $x$-intercepts are $p$ and $q$, the axis of symmetry is halfway between them, and the turning point lies on the axis.

   1. (a) $x - 1 = 0$ or $x - 5 = 0$, so $x = 1$ and $x = 5$
   2. (b) $x = 0$: $y = (-1)(-5) = 5$
   3. (c) $x = \frac{1 + 5}{2} = 3$
   4. (d) $y = (3 - 1)(3 - 5) = 2 \times (-2) = -4$, so $(3, -4)$

   Answer: (a) $1$ and $5$ (b) $5$ (c) $x = 3$ (d) $(3, -4)$

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 9 quadratics worksheet