Transformations of graphs
Algebra and functions
- A-level
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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The curve $y = x^2$ is translated by the vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Write down the equation of the new curve.
3 marks
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Describe the single transformation that maps the graph of $y = x^2$ onto the graph of $y = (x + 3)^2$.
3 marks
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Describe a single transformation that maps $y = 2^x$ onto $y = 2^{x - 1} + 3$.
4 marks
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Describe the transformations that map $y = \sin x$ onto $y = 3\sin x$, and the transformation that maps $y = \sin x$ onto $y = \sin 2x$.
4 marks
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The graph of $y = f(x)$ crosses the $x$-axis at $x = -1$, $x = 2$ and $x = 4$. Write down the $x$-coordinates of the points where $y = f(x - 3)$ crosses the $x$-axis, and where $y = f(2x)$ crosses the $x$-axis.
5 marks
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Find the coordinates of the points where the curve $y = x^2 - 4x + 3$ meets the line $y = 2x - 4$.
5 marks
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Solve $|2x - 4| = x + 1$ by considering the two linear pieces of the modulus graph.
6 marks
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The graph of $y = f(x)$, where $f(x) = x^3 - x$, is stretched by scale factor $2$ parallel to the $y$-axis and then translated by the vector $\begin{pmatrix} 0 \\ -3 \end{pmatrix}$. Find the equation of the resulting curve, and find the coordinates of its intersections with the line $y = -3$.
7 marks
Mark scheme
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Answer. $y = (x - 2)^2 - 1$
- A translation of $2$ in the $x$-direction replaces $x$ by $x - 2$.
- A translation of $-1$ in the $y$-direction subtracts $1$.
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Answer. A translation by the vector $\begin{pmatrix} -3 \\ 0 \end{pmatrix}$.
- $x$ has been replaced by $x + 3$.
- That is a translation of $3$ units in the negative $x$-direction.
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Answer. A translation by the vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$.
- $x$ is replaced by $x - 1$, which is a translation of $1$ in the positive $x$-direction.
- Adding $3$ translates the graph by $3$ in the positive $y$-direction.
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Answer. $y = 3\sin x$ is a stretch of scale factor $3$ parallel to the $y$-axis. $y = \sin 2x$ is a stretch of scale factor $\dfrac{1}{2}$ parallel to the $x$-axis.
- Multiplying the whole function by $3$ stretches distances from the $x$-axis by $3$.
- Replacing $x$ by $2x$ halves horizontal distances.
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Answer. $y = f(x - 3)$ crosses at $x = 2$, $x = 5$ and $x = 7$. $y = f(2x)$ crosses at $x = -\dfrac{1}{2}$, $x = 1$ and $x = 2$.
- A translation of $3$ in the positive $x$-direction adds $3$ to each intercept.
- Replacing $x$ by $2x$ halves each intercept.
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Answer. $\big(3 + \sqrt{2},\, 2 + 2\sqrt{2}\big)$ and $\big(3 - \sqrt{2},\, 2 - 2\sqrt{2}\big)$
- $x^2 - 4x + 3 = 2x - 4$.
- $x^2 - 6x + 7 = 0$, so $x = 3 \pm \sqrt{2}$.
- Substitute into $y = 2x - 4$.
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Answer. The solutions are $x = 1$ and $x = 5$.
- For $x \ge 2$, $2x - 4 = x + 1$, so $x = 5$, which is in this region.
- For $x < 2$, $4 - 2x = x + 1$, so $x = 1$, which is in this region.
- Both values satisfy the original equation.
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Answer. The equation is $y = 2(x^3 - x) - 3$. The intersections are $(-1, -3)$, $(0, -3)$ and $(1, -3)$.
- The stretch gives $y = 2(x^3 - x)$.
- The translation gives $y = 2(x^3 - x) - 3$.
- Set $2(x^3 - x) - 3 = -3$, so $x(x^2 - 1) = 0$.
- The solutions are $x = -1$, $x = 0$ and $x = 1$.