Year 10 Graphs of Parabolas, Circles, Hyperbolas and Exponentials Worksheets & Practice

Worked examples

Year 10 graphs of parabolas, circles, hyperbolas and exponentials: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Sketching a parabola from its equation

   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 $(0, 3)$.

2. Example 2 · Identifying and comparing curve types from equations

   For each equation below, state the type of curve, then write down one key feature.

   (a) $x^2 + y^2 = 25$ (b) $y = \dfrac{6}{x}$ (c) $y = 2^x$

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   Key idea: Match the algebraic form to the family: $x^2 + y^2 = r^2$ is a circle; $y = \tfrac{k}{x}$ is a hyperbola; $y = a^x$ ($a > 1$) is an exponential growth curve.

   1. (a) $x^2 + y^2 = 25$ matches $x^2 + y^2 = r^2$ with $r^2 = 25$, so $r = 5$. This is a circle centred at the origin with radius $5$.
   2. (b) $y = \dfrac{6}{x}$ matches $y = \dfrac{k}{x}$ with $k = 6 > 0$. This is a hyperbola with asymptotes along the $x$-axis ($y = 0$) and $y$-axis ($x = 0$), lying in the first and third quadrants.
   3. (c) $y = 2^x$ matches $y = a^x$ with $a = 2 > 1$. This is an exponential growth curve with horizontal asymptote $y = 0$ and $y$-intercept $(0, 1)$.

   Answer: (a) Circle, centre $(0, 0)$, radius $5$. (b) Hyperbola, asymptotes $x = 0$ and $y = 0$. (c) Exponential growth, horizontal asymptote $y = 0$.

3. Example 3 · Real-life exponential decay

   A scientist measures the mass of a radioactive sample. At time $t = 0$ the mass is $80$ grams. The mass halves every hour, so the mass in grams after $t$ hours is modelled by $M = 80 \times \left(\dfrac{1}{2}\right)^t$.

   (a) Write the mass when $t = 0$, $t = 1$, $t = 2$ and $t = 3$. (b) State the horizontal asymptote of the graph of $M$ against $t$ and explain what it means in this context. (c) After how many hours is the mass first below $15$ grams?

   Show worked solution

   Key idea: $M = 80 \times \left(\dfrac{1}{2}\right)^t$ is an exponential decay model. The base $\tfrac{1}{2} < 1$, so $M$ decreases and approaches $0$ but never reaches it.

   1. Evaluate at each time: $\begin{aligned} t=0:& M = 80 \times 1 = 80 \text{ g}\\ t=1:& M = 80 \times \tfrac{1}{2} = 40 \text{ g}\\ t=2:& M = 80 \times \tfrac{1}{4} = 20 \text{ g}\\ t=3:& M = 80 \times \tfrac{1}{8} = 10 \text{ g} \end{aligned}$
   2. The horizontal asymptote is $M = 0$. This means the mass gets closer and closer to zero grams but, according to the model, never actually reaches zero.
   3. From step 1, at $t = 2$ the mass is $20$ g (not yet below $15$ g), and at $t = 3$ the mass is $10$ g (below $15$ g). So the mass is first below $15$ grams after $3$ hours.

   Answer: (a) $80$ g, $40$ g, $20$ g, $10$ g. (b) Asymptote $M = 0$; the mass never reaches zero. (c) After $3$ hours.

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