Year 10 and 10A Parabolas and rates of change: free practice test with answers

Sample parabolas and rates of change questions with answers

Try each one before you open the answer.

1. Exploring parabolas

   For the parabola $y = x^{2} - 6x + 8$, find the value of $y$ when:

   (a) $x = 5$

   (b) $x = -1$

   Show the answer

   Key idea: Substitute the $x$ value into the rule.

   1. $x = 5$: $25 - 30 + 8 = 3$
   2. $x = -1$: $1 + 6 + 8 = 15$

   Answer: (a) $3$ (b) $15$

2. Sketching parabolas using transformations

   Write the coordinates of the turning point of $y = (x - 4)^{2} + 1$, in the form $(x, y)$.

   Show the answer

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

   1. Here $h = 4$ and $k = 1$

   Answer: $(4, 1)$

3. Sketching parabolas using factorisation

   Sketch the parabola $y = x^2 - 4x + 3$ and find its turning point, axis of symmetry, and intercepts.

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   Key idea: For $y = ax^2 + bx + c$, the axis of symmetry is $x = -\dfrac{b}{2a}$, and the turning point lies on this line.

   1. Find the axis of symmetry: $x = -\dfrac{-4}{2(1)} = 2$
   2. Find the turning point by substituting $x = 2$: $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$, so the turning point is $(2,\,-1)$.
   3. Find the $y$-intercept (set $x = 0$): $y = 0 - 0 + 3 = 3$, so the $y$-intercept is $(0, 3)$.
   4. Find the $x$-intercepts (set $y = 0$): $x^2 - 4x + 3 = 0 \Rightarrow (x-1)(x-3) = 0$, so $x = 1$ or $x = 3$; the $x$-intercepts are $(1, 0)$ and $(3, 0)$.
   5. The parabola opens upward ($a = 1 > 0$), has turning point $(2, -1)$, axis of symmetry $x = 2$, $y$-intercept $(0, 3)$, and $x$-intercepts $(1, 0)$ and $(3, 0)$.

   Answer: Turning point $(2, -1)$; axis of symmetry $x = 2$; $x$-intercepts $(1, 0)$ and $(3, 0)$; $y$-intercept $(0, 3)$.

4. Sketching parabolas by completing the square

   By completing the square, write $y = x^2 - 4x - 5$ in turning point form. Hence state the turning point, the $y$-intercept and the $x$-intercepts.

   Show the answer

   Key idea: Completing the square gives $y = (x - h)^2 + k$ with turning point $(h, k)$.

   1. $y = x^2 - 4x + 4 - 4 - 5 = (x - 2)^2 - 9$
   2. Turning point: $(2, -9)$ (a minimum, as the coefficient of $x^2$ is positive).
   3. $y$-intercept: let $x = 0$, then $y = -5$, giving $(0, -5)$.
   4. $x$-intercepts: $(x - 2)^2 = 9$, so $x - 2 = \pm 3$ and $x = -1$ or $x = 5$.

   Answer: $y = (x - 2)^2 - 9$; turning point $(2, -9)$; $y$-intercept $(0, -5)$; $x$-intercepts $(-1, 0)$ and $(5, 0)$

5. Applications of parabolas

   The entrance to a tunnel is a parabolic arch. With the origin at the left edge of the road, the arch is $y = -0.1(x - 10)^{2} + 10$, where $x$ and $y$ are in metres.

   (a) How wide is the arch at road level?

   (b) How high is the arch 4 m from the left edge?

   Show the answer

   Key idea: The arch meets the road where $y = 0$.

   1. $0.1(x - 10)^{2} = 10$ gives $(x - 10)^{2} = 100$, so $x = 0$ or $x = 20$; width $20$ m
   2. $y = -0.1(4 - 10)^{2} + 10 = -3.6 + 10 = 6.4$ m

   Answer: (a) $20$ m (b) $6.4$ m

6. Intersection of lines and parabolas

   Solve the simultaneous equations $y = x + 1$ and $y = x^2 - x - 2$.

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   Key idea: Substitute the linear expression for $y$ into the quadratic, then solve the resulting equation.

   1. Set the expressions equal: $x + 1 = x^2 - x - 2$
   2. Rearrange to standard form: $0 = x^2 - 2x - 3$
   3. Factorise: $0 = (x - 3)(x + 1) \implies x = 3 \text{ or } x = -1$
   4. Find $y$ for each $x$ using $y = x + 1$: $x = 3 \implies y = 4; x = -1 \implies y = 0$
   5. Check $(3, 4)$ in $y = x^2 - x - 2$: $9 - 3 - 2 = 4$ ✓. Check $(-1, 0)$: $1 + 1 - 2 = 0$ ✓

   Answer: $(3,\ 4)$ and $(-1,\ 0)$

7. Rates of change

   A cyclist rides $45$ km along a rail trail in $1.5$ hours. Find her average speed in km/h.

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   Key idea: Average speed is distance divided by time.

   1. $45 \div 1.5 = 30$ km/h

   Answer: $30$ km/h

8. Average and instantaneous rates of change

   Let $f(x) = x^{2}$. Find the average rate of change of $f$ from $x = 3$ to $x = 3.1$.

   Show the answer

   Key idea: Average rate of change $= \frac{f(3.1) - f(3)}{3.1 - 3}$.

   1. $f(3.1) = 9.61$ and $f(3) = 9$
   2. $\frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1$

   Answer: $6.1$

Free worksheets for this chapter

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

• Year 10 graphs of parabolas, circles, hyperbolas and exponentials worksheet
• Year 10 quadratic equations worksheet
• Year 10A factorising non-monic quadratics worksheet
• Year 10A linear and non-linear simultaneous equations worksheet
• Year 10 compound interest worksheet
• Year 10 direct and inverse proportion worksheet