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Year 12 Maths - Methods (Unit 3 & Unit 4) Transformations

Transformations of power functions

20 practice questions 0 video lessons Theory + worked examples
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Theory

A power function \(y=x^{n}\) — including the reciprocal \(\dfrac{1}{x}\), the reciprocal square \(\dfrac{1}{x^{2}}\) and the square root \(\sqrt{x}\) — can be dilated, reflected and translated into the form \(y=a(x-h)^{n}+k\). Here \(a\) is a dilation from the \(x\)-axis (with a reflection if \(a<0\)), \(h\) is a horizontal translation and \(k\) a vertical one, and from these you can sketch the curve and state its domain, range and asymptotes.

A power function has the form \(y=x^{n}\). As well as the polynomials \(x^{2},x^{3},x^{4},\dots\), this family includes the reciprocal \(\dfrac{1}{x}=x^{-1}\), the reciprocal square \(\dfrac{1}{x^{2}}=x^{-2}\), and the square root \(\sqrt{x}=x^{1/2}\). Each has a characteristic shape and a natural “base point” — the origin for \(x^{n}\), a pair of asymptotes for the reciprocals, and an endpoint for \(\sqrt{x}\).

Every such graph can be transformed by a combination of a dilation, a reflection and translations, giving the general rule \(y=a(x-h)^{n}+k\). The constant \(a\) stretches the graph by factor \(|a|\) from the \(x\)-axis (and reflects it in the \(x\)-axis when \(a<0\)); \(h\) slides it \(h\) units horizontally; and \(k\) slides it \(k\) units vertically. The reciprocal and root analogues are \(y=\dfrac{a}{x-h}+k\), \(y=\dfrac{a}{(x-h)^{2}}+k\) and \(y=a\sqrt{x-h}+k\).

The key features move with the graph. A turning or stationary point at the origin moves to \((h,k)\); the endpoint of a square-root curve moves to \((h,k)\) so its domain becomes \(x\ge h\); and the asymptotes of a reciprocal move to \(x=h\) and \(y=k\). Reading \(a,h,k\) off the rule lets you sketch the image and state its domain and range.

Key idea. Write the function as \(y=a(x-h)^{n}+k\) (or the reciprocal/root analogue). Then \(a\) dilates from the \(x\)-axis and reflects if negative, \((x-h)\) shifts right by \(h\), and \(+k\) shifts up by \(k\). Apply dilations and reflections before translations.
Cubic y=x cubed translated to y=(x-2) cubed + 1The curve y equals x cubed shown dashed, translated 2 right and 1 up to y equals (x minus 2) cubed plus 1; the point of zero gradient moves from the origin to (2,1) and the x-intercept is at (1,0). x y y=x^3
\(y=x^{3}\) (dashed) translated \(2\) right and \(1\) up to \(y=(x-2)^{3}+1\); the zero-gradient point moves to \((2,1)\)
Reciprocal y=1/x translated to y=1/(x-1)+2The curve y equals one over x shown dashed, translated 1 right and 2 up to y equals one over (x minus 1) plus 2; the vertical asymptote is x equals 1 and the horizontal asymptote is y equals 2, both shown as red dashed lines. x y y=2 x=1
\(y=\dfrac{1}{x}\) (dashed) translated \(1\) right and \(2\) up to \(y=\dfrac{1}{x-1}+2\); asymptotes move to \(x=1\) and \(y=2\)

The general transformed power function — \(a\) dilates from the \(x\)-axis (reflection if \(a<0\)), \(h\) and \(k\) translate:

\[y=a(x-h)^{n}+k\]
y=a(xh)n+k

The reciprocal and reciprocal-square analogues, with vertical asymptote \(x=h\) and horizontal asymptote \(y=k\):

\[y=\frac{a}{x-h}+k \qquad y=\frac{a}{(x-h)^{2}}+k\]
y=axh+k

The square-root analogue, with endpoint \((h,k)\) and domain \(x\ge h\):

\[y=a\sqrt{x-h}+k\]
y=axh+k
Reading off the transformations. \(a\): dilation by factor \(|a|\) from the \(x\)-axis, and a reflection in the \(x\)-axis if \(a<0\). \(h\): translation \(h\) units right. \(k\): translation \(k\) units up. Key point \((0,0)\to(h,k)\); reciprocal asymptotes \(\to x=h,\ y=k\).

How to sketch a transformed power function

  1. Write it in standard form. Express the rule as \(y=a(x-h)^{n}+k\) (or \(\dfrac{a}{x-h}+k\), \(\dfrac{a}{(x-h)^{2}}+k\), \(a\sqrt{x-h}+k\)) and read off \(a\), \(h\) and \(k\).
  2. Dilate and reflect first. Stretch the base graph by factor \(|a|\) from the \(x\)-axis; if \(a<0\), reflect it in the \(x\)-axis.
  3. Then translate. Move the graph \(h\) units horizontally (right if \(h>0\)) and \(k\) units vertically (up if \(k>0\)). Shift the key feature — the turning/stationary point, the endpoint, or the asymptotes — by \((h,k)\).
  4. Find intercepts. Set \(x=0\) for the \(y\)-intercept and \(y=0\) for the \(x\)-intercept(s), where they exist.
  5. State domain and range. Use the shifted asymptotes (reciprocals) or the endpoint and direction (square root) to write the domain and range.
Order matters. Apply the dilation/reflection (the \(a\)) before the translations (the \(h\) and \(k\)). Reversing the order can leave the graph in the wrong place.
Example 1 — translated cubic
Sketch \(y=(x-2)^{3}+1\). State the point of zero gradient and the axis intercepts.
Solution
Compare with \(y=(x-h)^{3}+k\) — read off the translation:
\(h,\ k\)\(=\)\(2,\ 1\)
The stationary point of inflection of \(y=x^{3}\) is at \((0,0)\), so it moves to:
\((0,0)\)\(\to\)\((2,1)\)
\(x\)-intercept — set \(y=0\):
\((x-2)^{3}+1\)\(=\)\(0\)
\((x-2)^{3}\)\(=\)\(-1\)
\(x-2\)\(=\)\(-1\)
\(x\)\(=\)\(1\)
\(y\)-intercept — set \(x=0\):
\(y\)\(=\)\((-2)^{3}+1=-7\)
\(\therefore\) zero gradient at \((2,1)\); \(x\)-intercept \((1,0)\); \(y\)-intercept \((0,-7)\)
(2,1)
Example 2 — reciprocal: asymptotes, domain, range
For \(y=\dfrac{2}{x+1}-3\), state the transformations of \(y=\dfrac{1}{x}\), the asymptotes, the domain and the range.
Solution
Compare with \(y=\dfrac{a}{x-h}+k\) — note \(x+1=x-(-1)\):
\(a,\ h,\ k\)\(=\)\(2,\ -1,\ -3\)
Transformations of \(y=\dfrac{1}{x}\):
dilationfactor \(2\) from the \(x\)-axis
translation\(1\) left, \(3\) down
Asymptotes move from \(x=0,\ y=0\) to \(x=h,\ y=k\):
vertical\(:\)\(x=-1\)
horizontal\(:\)\(y=-3\)
\(\therefore\) domain \(x\in\mathbb{R}\setminus\{-1\}\); range \(y\in\mathbb{R}\setminus\{-3\}\)
x=1,y=3
Example 3 — reflected square root
Sketch \(y=2-\sqrt{x+3}\). State the transformations of \(y=\sqrt{x}\), the domain, range and axis intercepts.
Solution
Write as \(y=a\sqrt{x-h}+k\) — here \(y=-\sqrt{x-(-3)}+2\):
\(a,\ h,\ k\)\(=\)\(-1,\ -3,\ 2\)
So \(y=\sqrt{x}\) is reflected in the \(x\)-axis (\(a<0\)), then translated \(3\) left and \(2\) up; the endpoint moves to \((h,k)\):
\((0,0)\)\(\to\)\((-3,2)\)
Domain — need \(x+3\ge 0\); range — reflected, so \(y\le k\):
domain\(:\)\(x\ge -3\)
range\(:\)\(y\le 2\)
\(x\)-intercept — set \(y=0\):
\(\sqrt{x+3}\)\(=\)\(2\)
\(x+3\)\(=\)\(4\)
\(x\)\(=\)\(1\)
\(\therefore\) endpoint \((-3,2)\); domain \(x\ge -3\); range \(y\le 2\); \(x\)-intercept \((1,0)\), \(y\)-intercept \((0,2-\sqrt{3})\)
Square root y=sqrt(x) reflected and translated to y=-sqrt(x+3)+2The transformed square-root curve y equals minus root (x plus 3) plus 2 starts at the endpoint (-3,2) and decreases to the right, crossing the x-axis at (1,0); the domain is x greater than or equal to -3 and the range is y less than or equal to 2. x y (-3,2) (1,0)
y2
Example 4 — sequence of transformations
Describe a sequence of transformations that maps \(y=x^{3}\) onto \(y=5-2(x+4)^{3}\).
Solution
Rewrite in standard form \(y=a(x-h)^{3}+k\):
\(y\)\(=\)\(-2(x+4)^{3}+5\)
\(a,\ h,\ k\)\(=\)\(-2,\ -4,\ 5\)
Interpret \(a=-2\) — dilation and reflection (applied first):
\(|a|=2\)\(\Rightarrow\)dilation factor \(2\) from the \(x\)-axis
\(a<0\)\(\Rightarrow\)reflection in the \(x\)-axis
Interpret \(h=-4,\ k=5\) — translations (applied last):
\(h=-4\)\(\Rightarrow\)translate \(4\) left
\(k=5\)\(\Rightarrow\)translate \(5\) up
\(\therefore\) dilate by factor \(2\) from the \(x\)-axis, reflect in the \(x\)-axis, then translate \(4\) left and \(5\) up
y=2(x+4)3+5

Common pitfalls

Watch the direction of the horizontal shift. \((x-h)\) moves the graph right by \(h\). So \(y=(x-2)^{3}\) is \(y=x^{3}\) shifted \(2\) right, and \(y=(x+2)^{3}\) is shifted \(2\) left — opposite to the sign inside the bracket.
\(a\) dilates from the \(x\)-axis, not the \(y\)-axis. In \(y=a(x-h)^{n}+k\) the factor \(a\) multiplies the whole output, stretching vertically by \(|a|\); a negative \(a\) reflects in the \(x\)-axis and flips the range.
Move the asymptotes and re-state the domain. After \(y=\dfrac{a}{x-h}+k\) the asymptotes are \(x=h\) and \(y=k\), not \(x=0\) and \(y=0\); the domain excludes \(x=h\) and the range excludes \(y=k\).

Frequently asked questions

What do \(a\), \(h\) and \(k\) do in \(y=a(x-h)^{n}+k\)?

\(a\) is a dilation by factor \(|a|\) from the \(x\)-axis, and reflects in the \(x\)-axis if \(a<0\); \(h\) is a horizontal translation (right if \(h>0\)); and \(k\) is a vertical translation (up if \(k>0\)). Together they map \(y=x^{n}\) onto the transformed curve.

Which way does \(y=(x-h)^{n}\) move compared with \(y=x^{n}\)?

It moves horizontally by \(h\), opposite to the sign inside the bracket. So \(y=(x-2)^{3}\) is \(y=x^{3}\) shifted \(2\) right, while \(y=(x+2)^{3}\) is shifted \(2\) left.

How do you find the asymptotes of a transformed reciprocal function?

For \(y=\dfrac{a}{x-h}+k\), the vertical asymptote is \(x=h\) (denominator zero) and the horizontal asymptote is \(y=k\). The same holds for the reciprocal square \(y=\dfrac{a}{(x-h)^{2}}+k\).

What are the domain and range of \(y=a\sqrt{x-h}+k\)?

The root needs \(x-h\ge 0\), so the domain is \(x\ge h\) and the curve starts at \((h,k)\). If \(a>0\) the range is \(y\ge k\); if \(a<0\) it is reflected, so \(y\le k\).

In what order do you apply the transformations?

Dilations and reflections first, then translations. For \(y=a(x-h)^{n}+k\), scale by \(|a|\) and reflect if \(a<0\), then translate by \(h\) and \(k\). Translating first can place the graph wrongly.

How do you sketch a transformed power function?

Read \(a\), \(h\), \(k\) off the rule, move the key feature (turning point, endpoint or asymptotes) by \((h,k)\), apply the dilation and any reflection, find the intercepts, and state the domain and range.