Special graphs
Algebra
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Write the equation of the circle with centre $(0, 0)$ and radius $5$.
2 marks
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On $y = \dfrac{1}{x}$, what is $y$ when $x = 2$? What happens to $y$ as $x$ gets very close to $0$?
2 marks
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Complete the table of $y = 2^{x}$ for $x = 0, 1, 2, 3$.
3 marks
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For $y = x^{3} + 1$, find $y$ when $x = -1$ and when $x = 2$.
3 marks
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The circle $x^{2} + y^{2} = 100$ has radius $10$. Does $(6, 8)$ lie on it? Does $(5, 5)$ lie on it?
4 marks
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For $y = 2^{x}$, find $y$ when $x = 0$ and when $x = -2$. Solve $2^{x} = 16$, and write the $y$-intercept.
4 marks
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Describe the graph of $y = -x^{2} + 4$, including the intercepts and the turning point. For $y = \dfrac{1}{x - 1}$, write the equations of the asymptotes and the value of $y$ when $x = 2$.
5 marks
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A circle has centre $(0, 0)$ and passes through $(3, 4)$. Write its equation and the points where it meets the $y$-axis. Then find where the line $y = x - 1$ meets the circle.
6 marks
Mark scheme
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Answer. $x^{2} + y^{2} = 25$
- The radius squared is $25$.
- $x^{2} + y^{2} = 25$
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Answer. $y = \dfrac{1}{2}$, and $y$ becomes very large.
- $\dfrac{1}{2}$
- The $y$-axis is an asymptote, so $y$ increases or decreases without bound.
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Answer. $1, 2, 4, 8$
- $2^{0} = 1$ and $2^{1} = 2$.
- $2^{2} = 4$ and $2^{3} = 8$.
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Answer. $0$ and $9$
- $(-1)^{3} + 1 = 0$
- $8 + 1 = 9$
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Answer. Yes, and no
- $36 + 64 = 100$, so $(6, 8)$ is on the circle.
- $25 + 25 = 50 \ne 100$, so $(5, 5)$ is not.
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Answer. $1$, $\dfrac{1}{4}$, $x = 4$, and $(0, 1)$
- $2^{0} = 1$ and $2^{-2} = \dfrac{1}{4}$.
- $2^{4} = 16$, so $x = 4$.
- The curve crosses the $y$-axis at $(0, 1)$.
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Answer. Maximum $(0, 4)$, $x$-intercepts $(\pm 2, 0)$. Asymptotes $x = 1$ and $y = 0$. When $x = 2$, $y = 1$.
- The quadratic is a downward parabola with maximum $(0, 4)$.
- It crosses the $x$-axis where $x^{2} = 4$.
- The reciprocal graph has asymptotes $x = 1$ and $y = 0$.
- When $x = 2$, $y = 1$.
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Answer. Equation $x^{2} + y^{2} = 25$. It meets the $y$-axis at $(0, 5)$ and $(0, -5)$. The line meets it at $(4, 3)$ and $(-3, -4)$.
- $3^{2} + 4^{2} = 25$, so the equation is $x^{2} + y^{2} = 25$.
- When $x = 0$, $y = \pm 5$.
- $x^{2} + (x - 1)^{2} = 25$ gives $x^{2} - x - 12 = 0$.
- The points are $(4, 3)$ and $(-3, -4)$.