Resources For Teachers For Tutors For Students & Parents Pricing
Year 12 Maths - Methods (Unit 3 & Unit 4) Transformations

Using transformations to sketch graphs

20 practice questions 0 video lessons Theory + worked examples
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

Sketching a transformed graph starts from a known base curve — such as \(y=x^{2}\), \(y=\dfrac{1}{x}\), \(y=\sqrt{x}\), \(y=\sin x\) or \(y=2^{x}\) — and applies dilations, reflections and translations written together as \(y=a\,f\big(b(x-h)\big)+k\). You map the base graph's key points and asymptotes to their images, then draw the result with its intercepts, asymptotes, domain and range labelled.

A base graph is a standard curve whose shape you already know: the parabola \(y=x^{2}\), the hyperbola \(y=\dfrac{1}{x}\), the square-root curve \(y=\sqrt{x}\), the exponential \(y=2^{x}\) and the trigonometric curve \(y=\sin x\). A transformation reshapes or repositions this base graph without changing which family it belongs to, so the transformed curve keeps the same essential shape.

There are three families of transformation. A dilation stretches or compresses the graph — a factor \(a\) outside the function stretches it vertically, and a factor \(b\) inside compresses it horizontally. A reflection flips the graph in an axis: \(-f(x)\) flips it in the \(x\)-axis and \(f(-x)\) flips it in the \(y\)-axis. A translation slides the graph, by \(h\) horizontally and \(k\) vertically. Collected together, a transformed standard curve is \(y=a\,f\big(b(x-h)\big)+k\).

To sketch it, you do not plot dozens of points. Instead you map the key features of the base graph — its key points, intercepts, endpoints and any asymptotes — to their new positions, plot those images, and draw the curve through them with the correct shape and end behaviour. Finally you read off and state the domain and range, which the dilations, reflections and translations may have changed.

Key idea. Write the rule as \(y=a\,f\big(b(x-h)\big)+k\), then send each base point \((x,y)\) to \(\left(\dfrac{x}{b}+h,\;a\,y+k\right)\). Asymptotes and endpoints move with the graph, so redraw them in their new positions before sketching.
Translating a parabolaThe base parabola y=x squared shown dashed with vertex at the origin, and its image y=(x-2) squared plus 1 shown solid with vertex translated to (2,1), moved right 2 and up 1. x y y=x^2 y=(x-2)^2+1
Translate \(y=x^{2}\) (dashed) right \(2\) and up \(1\): the vertex \((0,0)\) maps to \((2,1)\)
Transformed hyperbola with asymptotesThe curve y=1/(x-1)+2, a hyperbola translated right 1 and up 2, with a vertical asymptote at x=1 and a horizontal asymptote at y=2 shown dashed in red. x y y=1/(x-1)+2 x=1 y=2
\(y=\dfrac{1}{x-1}+2\): translating \(y=\dfrac{1}{x}\) moves the asymptotes to \(x=1\) and \(y=2\)

A transformed standard curve, with base function \(f\):

\[y=a\,f\big(b(x-h)\big)+k\]
y=af(b(xh))+k

Each base point \((x,y)\) is sent to its image by the point-mapping rule:

\[(x,y)\;\longmapsto\;\left(\dfrac{x}{b}+h,\;a\,y+k\right)\]
(x,y)(xb+h,ay+k)

The individual effects — a vertical dilation and \(x\)-axis reflection from \(a\), a horizontal dilation and \(y\)-axis reflection from \(b\), and the translations from \(h,k\):

\[\text{vert. dilation }|a|,\quad \text{horiz. dilation }\dfrac{1}{|b|},\quad \text{right }h,\quad \text{up }k\]
|a|,1|b|,h,k
Reading the signs. If \(a<0\) the graph is reflected in the \(x\)-axis; if \(b<0\) it is reflected in the \(y\)-axis. A horizontal asymptote \(y=0\) becomes \(y=k\), and a vertical asymptote \(x=0\) becomes \(x=h\).

How to sketch a transformed graph

  1. Name the base graph. Recognise which standard curve \(f\) you are transforming — \(x^{2}\), \(\dfrac{1}{x}\), \(\sqrt{x}\), \(\sin x\) or \(a^{x}\) — and recall its shape and key features.
  2. Write it in standard form. Rearrange the rule as \(y=a\,f\big(b(x-h)\big)+k\); factorise the \(b\) out of the bracket so the horizontal shift \(h\) is read correctly.
  3. Read off the transformations. The dilation factors are \(|a|\) (vertical) and \(\dfrac{1}{|b|}\) (horizontal); negative \(a\) or \(b\) gives a reflection; \(h\) and \(k\) are the translations.
  4. Map the key features. Send each key point \((x,y)\) to \(\left(\dfrac{x}{b}+h,\,a\,y+k\right)\), and move every asymptote and endpoint to its new position.
  5. Draw and label. Plot the images, draw the curve with the correct shape and end behaviour, mark the intercepts and asymptotes, and state the domain and range.
Order tip. Apply dilations and reflections before translations. Better still, use the point-mapping rule \(\left(\dfrac{x}{b}+h,\,a\,y+k\right)\) directly — it builds the order in automatically, so you never shift by the wrong amount.
Example 1 — translation of a parabola
Sketch \(y=(x-2)^{2}+1\) from the base graph \(y=x^{2}\), showing the vertex and \(y\)-intercept, and state the domain and range.
Solution
Read off the translation — \(h=2\) (right), \(k=1\) (up); no dilation or reflection:
vertex \((0,0)\)\(\mapsto\)\((0+2,\;0+1)=(2,1)\)
Find the \(y\)-intercept — put \(x=0\):
\(y\)\(=\)\((0-2)^{2}+1\)
\(=\)\(5\)
The parabola opens up with vertex \((2,1)\), so it never falls below \(y=1\):
\(\therefore\) vertex \((2,1)\), \(y\)-intercept \((0,5)\); domain \(x\in\mathbb{R}\), range \(y\ge 1\)
y=(x2)2+1
Example 2 — hyperbola with moved asymptotes
Sketch \(y=\dfrac{1}{x-1}+2\) from \(y=\dfrac{1}{x}\), giving the asymptotes and intercepts, and state the domain and range.
Solution
Read off the translation — \(h=1\), \(k=2\); the asymptotes move with the graph:
\(x=0\)\(\mapsto\)\(x=1\) (vertical)
\(y=0\)\(\mapsto\)\(y=2\) (horizontal)
\(y\)-intercept — put \(x=0\); \(x\)-intercept — put \(y=0\):
\(x=0:\;y\)\(=\)\(\dfrac{1}{-1}+2=1\)
\(y=0:\;\dfrac{1}{x-1}\)\(=\)\(-2\;\Rightarrow\;x=\dfrac{1}{2}\)
\(\therefore\) asymptotes \(x=1,\;y=2\); intercepts \(\left(0,1\right),\left(\dfrac12,0\right)\); domain \(x\ne 1\), range \(y\ne 2\)
y=1x1+2
Example 3 — reflection and dilation of \(\sqrt{x}\)
Sketch \(y=3-2\sqrt{x}\) from \(y=\sqrt{x}\). Identify the transformations, the endpoint and \(x\)-intercept, and state the domain and range.
Solution
Write in standard form — \(y=-2\sqrt{x}+3\), so \(a=-2\) (reflect in \(x\)-axis, dilate by \(2\)), \(k=3\):
endpoint \((0,0)\)\(\mapsto\)\((0,\;-2(0)+3)=(0,3)\)
\((1,1)\)\(\mapsto\)\((1,\;-2(1)+3)=(1,1)\)
\(x\)-intercept — put \(y=0\):
\(3-2\sqrt{x}\)\(=\)\(0\)
\(\sqrt{x}\)\(=\)\(\dfrac{3}{2}\;\Rightarrow\;x=\dfrac{9}{4}\)
\(\therefore\) starts at \((0,3)\) and falls; \(x\)-intercept \(\left(\dfrac94,0\right)\); domain \(x\ge 0\), range \(y\le 3\)
Reflected and dilated square-root curveThe base curve y=root x shown dashed rising from the origin, and its image y=3 minus 2 root x shown solid, reflected in the x-axis, stretched by factor 2 and translated up 3 so it starts at (0,3) and falls. x y y=√x y=3-2√x
y=32x
Example 4 — dilations of a sine curve
Sketch one period of \(y=2\sin(2x)+1\) from \(y=\sin x\). State the amplitude, period and range, and the coordinates of the maximum and minimum in \(0\le x\le\pi\).
Solution
Read off the dilations and translation — \(a=2\), \(b=2\), \(k=1\):
amplitude\(=\)\(|a|=2\)
period\(=\)\(\dfrac{2\pi}{b}=\dfrac{2\pi}{2}=\pi\)
midline\(=\)\(y=1\)
Range — the midline \(\pm\) amplitude:
range\(=\)\([\,1-2,\;1+2\,]=[-1,3]\)
Key points — maximum a quarter-period in, minimum three-quarters in:
max\(=\)\(\left(\dfrac{\pi}{4},\,3\right)\)
min\(=\)\(\left(\dfrac{3\pi}{4},\,-1\right)\)
\(\therefore\) amplitude \(2\), period \(\pi\), range \([-1,3]\); max \(\left(\dfrac{\pi}{4},3\right)\), min \(\left(\dfrac{3\pi}{4},-1\right)\)
y=2sin(2x)+1

Common pitfalls

Translations go the opposite way to the sign inside. \(y=f(x-2)\) shifts the graph \(2\) units right, not left; \(y=f(x+2)\) shifts it left. Only the outside constant \(k\) moves the graph in the direction of its sign.
Factor out \(b\) before reading the shift. In \(y=f(2x-4)\) the shift is not \(4\). Write \(2x-4=2(x-2)\), so the graph is compressed by \(\dfrac12\) and translated right \(2\), not right \(4\).
Move the asymptotes with the curve. For \(y=\dfrac{1}{x-1}+2\) the asymptotes are \(x=1\) and \(y=2\), not \(x=0\) and \(y=0\). Forgetting to shift them is the most common hyperbola error, and it also changes the range.

Frequently asked questions

What are the three main types of graph transformation?

Dilations stretch or compress a graph, reflections flip it in the \(x\)- or \(y\)-axis, and translations slide it left, right, up or down. Together they give \(y=a\,f\big(b(x-h)\big)+k\), where \(a,b\) set the dilations and reflections and \(h,k\) the translation.

Which way does the graph of y equals f of (x minus h) move?

To the right by \(h\). A change inside the function acts opposite to its sign, so \(f(x-2)\) shifts \(2\) right and \(f(x+2)\) shifts \(2\) left; adding \(k\) outside moves the graph straight up by \(k\).

How do dilations change a graph?

The outside factor \(a\) multiplies every height, stretching the graph vertically by \(|a|\) (and reflecting in the \(x\)-axis if \(a<0\)). The inside factor \(b\) dilates horizontally by \(\dfrac{1}{|b|}\); for example \(y=\sin(bx)\) has period \(\dfrac{2\pi}{b}\).

How do you sketch a transformed graph step by step?

Write the rule as \(y=a\,f\big(b(x-h)\big)+k\) and read off the dilations, reflections and translations. Then map the base graph's key points, intercepts, endpoints and asymptotes to their images, plot them, draw the curve, and label the domain and range.

How do transformations affect asymptotes?

Asymptotes move with the curve. For \(y=\dfrac{1}{x-1}+2\) the vertical asymptote of \(y=\dfrac{1}{x}\) moves to \(x=1\) and the horizontal one to \(y=2\). Always redraw them in their new positions before sketching the branches.

What is the general form of a transformed function?

It is \(y=a\,f\big(b(x-h)\big)+k\): \(a\) is the vertical dilation (its sign an \(x\)-axis reflection), \(b\) gives the horizontal dilation \(\dfrac{1}{|b|}\) (its sign a \(y\)-axis reflection), and \(h,k\) translate right \(h\) and up \(k\).