How do you find the image of a point under these transformations?

Apply the mapping rule to the coordinates. For y=a f(x) use (x,y) to (x, a y); for y=f(nx) use (x,y) to (x over n, y); for y=-f(x) use (x,y) to (x, -y); for y=f(-x) use (x,y) to (-x, y). Transform the key points, intercepts and asymptotes this way, then sketch the image.

How do you find the image of a point under these transformations?

Apply the mapping rule to the coordinates. For y=a f(x) use (x,y) to (x, a y); for y=f(nx) use (x,y) to (x over n, y); for y=-f(x) use (x,y) to (x, -y); for y=f(-x) use (x,y) to (-x, y). Transform the key points, intercepts and asymptotes this way, then sketch the image.

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

Dilations and reflections

Q: What does y=a f(x) do to a graph?

y=a f(x) is a dilation from the x-axis by factor a: every y-coordinate is multiplied by a, so the point (x,y) maps to (x, a y). Points on the x-axis stay fixed, so the x-intercepts do not move, while the range is multiplied by a. If a is negative the graph is also reflected in the x-axis.

Q: What is a dilation from the y-axis?

y=f(nx) is a dilation from the y-axis by factor one over n: the x-coordinate of every point is divided by n, so (x,y) maps to (x over n, y). The y-intercept does not move, the x-intercepts and any vertical asymptotes are divided by n, and the range is unchanged. For example y=f(2x) squashes the graph horizontally to half its width.

Q: How do you reflect a graph in the x-axis or the y-axis?

To reflect in the x-axis use y=-f(x): change the sign of every y-coordinate, so (x,y) maps to (x, -y). To reflect in the y-axis use y=f(-x): change the sign of every x-coordinate, so (x,y) maps to (-x, y). It is easy to swap them, so remember that the minus outside the function acts on y and the minus inside acts on x.

Q: Does a dilation change the x-intercepts?

A dilation from the x-axis, y=a f(x) with a positive, leaves the x-intercepts fixed because a times 0 is still 0. A dilation from the y-axis, y=f(nx), divides every x-intercept by n. So y=3f(x) keeps the same x-intercepts, but y=f(3x) moves them to one third of their original x-values.

Q: What happens to asymptotes under a dilation or reflection?

A horizontal asymptote y=k is multiplied by a under y=a f(x), becoming y=a k, and it changes sign under a reflection in the x-axis. A vertical asymptote x=c is divided by n under y=f(nx) and changes sign under a reflection in the y-axis. An asymptote at y=0 or x=0 stays where it is.

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A dilation stretches a graph away from an axis and a reflection flips it across an axis. \(y=a\,f(x)\) is a dilation from the \(x\)-axis by factor \(a\), \(y=f(nx)\) is a dilation from the \(y\)-axis by factor \(\dfrac{1}{n}\), while \(y=-f(x)\) reflects in the \(x\)-axis and \(y=f(-x)\) reflects in the \(y\)-axis. Track a few key points, intercepts and asymptotes and the image follows.

A dilation from the \(x\)-axis is produced by \(y=a\,f(x)\): every \(y\)-coordinate is multiplied by \(a\), so the point \((x,y)\) maps to \((x,\,a y)\). This is a dilation from the \(x\)-axis by factor \(a\). Points already on the \(x\)-axis stay put, so the \(x\)-intercepts do not move; the range is multiplied by \(a\) while the domain is unchanged. If \(a<0\) the graph is also reflected in the \(x\)-axis.

A dilation from the \(y\)-axis is produced by \(y=f(nx)\): replacing \(x\) with \(nx\) divides every \(x\)-coordinate by \(n\), so \((x,y)\) maps to \(\left(\dfrac{x}{n},\,y\right)\). This is a dilation from the \(y\)-axis by factor \(\dfrac{1}{n}\). The \(y\)-intercept is fixed, the \(x\)-intercepts and any vertical asymptotes are divided by \(n\), the range is unchanged and the domain is scaled.

A reflection in the \(x\)-axis is \(y=-f(x)\), which changes the sign of every \(y\)-coordinate: \((x,y)\mapsto(x,\,-y)\). A reflection in the \(y\)-axis is \(y=f(-x)\), which changes the sign of every \(x\)-coordinate: \((x,y)\mapsto(-x,\,y)\). A horizontal asymptote \(y=k\) becomes \(y=-k\) under the first; a vertical asymptote \(x=c\) becomes \(x=-c\) under the second.

Key idea. The number outside the function acts on \(y\): \(y=a\,f(x)\) gives \((x,y)\mapsto(x,ay)\). The number inside acts on \(x\): \(y=f(nx)\) gives \((x,y)\mapsto\left(\dfrac{x}{n},y\right)\). Reflections just insert a minus sign — outside for the \(x\)-axis, inside for the \(y\)-axis.

\(y=\dfrac{1}{2}f(x)\) is a dilation of \(y=f(x)\) from the \(x\)-axis by factor \(\dfrac{1}{2}\): the \(x\)-intercepts stay at \(\pm 2\), heights halve

\(y=-f(x)\) is the reflection of \(y=f(x)\) in the \(x\)-axis: each \(y\)-coordinate changes sign, the \(x\)-intercepts \(0\) and \(2\) are fixed

Dilation from the \(x\)-axis by factor \(a\) — multiply every \(y\)-coordinate by \(a\):

\[y=a\,f(x):\qquad (x,y)\mapsto(x,\,ay)\]

(x, y) \mapsto (x, a y)

Dilation from the \(y\)-axis by factor \(\dfrac{1}{n}\) — divide every \(x\)-coordinate by \(n\):

\[y=f(nx):\qquad (x,y)\mapsto\left(\dfrac{x}{n},\,y\right)\]

(x, y) \mapsto (\frac{x}{n}, y)

Reflection in the \(x\)-axis (negate \(y\)) and reflection in the \(y\)-axis (negate \(x\)):

\[y=-f(x):\ (x,y)\mapsto(x,\,-y) \qquad y=f(-x):\ (x,y)\mapsto(-x,\,y)\]

(x, y) \mapsto (x, - y); (x, y) \mapsto (- x, y)

Intercepts and asymptotes. Under \(y=a\,f(x)\) the \(x\)-intercepts are fixed and a horizontal asymptote \(y=k\) becomes \(y=ak\). Under \(y=f(nx)\) the \(x\)-intercepts and vertical asymptotes are divided by \(n\) while the \(y\)-intercept is fixed.

How to apply a dilation or reflection

Name the transformation. Decide which form you have: \(a\,f(x)\) (outside → acts on \(y\)), \(f(nx)\) (inside → acts on \(x\)), \(-f(x)\) or \(f(-x)\) (a reflection).
Read off the factor and axis. \(a\,f(x)\) is a dilation from the \(x\)-axis by factor \(a\); \(f(nx)\) is a dilation from the \(y\)-axis by factor \(\dfrac{1}{n}\); the minus sign gives the reflection axis.
Map the key features. Apply the rule to each turning point, intercept and asymptote — \((x,y)\mapsto(x,ay)\), or \(\left(\dfrac{x}{n},y\right)\), or a sign change on \(x\) or \(y\).
Update domain and range. A dilation from the \(x\)-axis scales the range; a dilation from the \(y\)-axis scales the domain; a reflection swaps the sign of one of them.
Sketch. Plot the transformed key points and asymptotes, then draw the image with the same overall shape.

Quick check. A change outside \(f\) moves points vertically (a \(y\)-effect); a change inside \(f\) moves points horizontally (an \(x\)-effect) — and the horizontal effect is always the reciprocal, so \(f(nx)\) shrinks widths by factor \(n\).

Example 1 — dilations, point mapping

The point \((6,4)\) lies on \(y=f(x)\). Find its image on (a) \(y=2f(x)\) and (b) \(y=f(3x)\).

Solution

(a) \(y=2f(x)\)

Dilation from the \(x\)-axis by factor \(2\) — multiply \(y\) by \(2\):

\((x,y)\)	\(\mapsto\)	\((x,\,2y)\)
\((6,4)\)	\(\mapsto\)	\((6,\,8)\)

\(\therefore\) image is \((6,8)\)

(b) \(y=f(3x)\)

Dilation from the \(y\)-axis by factor \(\dfrac{1}{3}\) — divide \(x\) by \(3\):

\((x,y)\)	\(\mapsto\)	\(\left(\dfrac{x}{3},\,y\right)\)
\((6,4)\)	\(\mapsto\)	\((2,\,4)\)

\(\therefore\) image is \((2,4)\)

(6, 8); (2, 4)

Example 2 — reflections, point mapping

The point \((-3,5)\) lies on \(y=f(x)\). Find its image on (a) \(y=-f(x)\) and (b) \(y=f(-x)\), and name each transformation.

Solution

(a) \(y=-f(x)\)

Reflection in the \(x\)-axis — change the sign of \(y\):

\((x,y)\)	\(\mapsto\)	\((x,\,-y)\)
\((-3,5)\)	\(\mapsto\)	\((-3,\,-5)\)

\(\therefore\) \((-3,-5)\); reflection in the \(x\)-axis

(b) \(y=f(-x)\)

Reflection in the \(y\)-axis — change the sign of \(x\):

\((x,y)\)	\(\mapsto\)	\((-x,\,y)\)
\((-3,5)\)	\(\mapsto\)	\((3,\,5)\)

\(\therefore\) \((3,5)\); reflection in the \(y\)-axis

(- 3, - 5); (3, 5)

Example 3 — asymptote and intercept

The graph of \(y=f(x)\) has a horizontal asymptote \(y=2\) and cuts the \(y\)-axis at \((0,5)\). State the asymptote and \(y\)-intercept of (a) \(y=\dfrac{1}{2}f(x)\) and (b) \(y=f(-x)\).

Solution

(a) \(y=\dfrac{1}{2}f(x)\)

Dilation from the \(x\)-axis by factor \(\dfrac{1}{2}\) — halve every \(y\)-value, including the asymptote:

asymptote	\(:\)	\(y=\dfrac{1}{2}\times 2=1\)
\((0,5)\)	\(\mapsto\)	\(\left(0,\,\dfrac{5}{2}\right)\)

\(\therefore\) asymptote \(y=1\); \(y\)-intercept \(\left(0,\dfrac{5}{2}\right)\)

(b) \(y=f(-x)\)

Reflection in the \(y\)-axis — \(y\)-values are unchanged and \(x=0\) maps to \(x=0\):

asymptote	\(:\)	\(y=2\) (unchanged)
\((0,5)\)	\(\mapsto\)	\((0,\,5)\)

\(\therefore\) asymptote \(y=2\); \(y\)-intercept \((0,5)\)

y = 1; (0, \frac{5}{2})

Example 4 — transform a rule and sketch

Given \(f(x)=4-x^{2}\), find the rule for \(y=f(2x)\), state its \(x\)-intercepts and turning point, and sketch it against \(y=f(x)\).

Solution

Substitute \(2x\) for \(x\) — this is a dilation from the \(y\)-axis by factor \(\dfrac{1}{2}\):

\(f(2x)\)	\(=\)	\(4-(2x)^{2}\)
	\(=\)	\(4-4x^{2}\)

\(x\)-intercepts — set \(4-4x^{2}=0\):

\(x^{2}\)	\(=\)	\(1\)
\(x\)	\(=\)	\(\pm 1\)

The maximum \((0,4)\) is on the \(y\)-axis, so it is unchanged; the intercepts move from \(\pm 2\) to \(\pm 1\).

\(\therefore\) \(y=4-4x^{2}\); \(x\)-intercepts \((\pm 1,0)\); maximum \((0,4)\)

y = 4 - 4 x^{2}

Common pitfalls

\(y=f(nx)\) dilates by \(\dfrac{1}{n}\), not \(n\). The \(x\)-coordinates are divided by \(n\), so \(y=f(3x)\) squashes the graph to one third of its width — do not stretch it three times wider.

Outside acts on \(y\), inside acts on \(x\). \(y=a\,f(x)\) changes \(y\)-coordinates (dilation from the \(x\)-axis); \(y=f(nx)\) changes \(x\)-coordinates (dilation from the \(y\)-axis). Do not apply the factor to the wrong coordinate.

Do not swap the two reflections. \(y=-f(x)\) (minus outside) is a reflection in the \(x\)-axis, negating \(y\); \(y=f(-x)\) (minus inside) is a reflection in the \(y\)-axis, negating \(x\).

Frequently asked questions

What does y=a f(x) do to a graph?

It is a dilation from the \(x\)-axis by factor \(a\): every \(y\)-coordinate is multiplied by \(a\), so \((x,y)\mapsto(x,ay)\). The \(x\)-intercepts stay fixed and the range is multiplied by \(a\); if \(a<0\) the graph is also reflected in the \(x\)-axis.

What is a dilation from the y-axis?

\(y=f(nx)\) is a dilation from the \(y\)-axis by factor \(\dfrac{1}{n}\): every \(x\)-coordinate is divided by \(n\), so \((x,y)\mapsto\left(\dfrac{x}{n},y\right)\). The \(y\)-intercept is fixed and \(x\)-intercepts and vertical asymptotes are divided by \(n\).

How do you reflect a graph in the x-axis or the y-axis?

Use \(y=-f(x)\) to reflect in the \(x\)-axis (negate \(y\): \((x,y)\mapsto(x,-y)\)) and \(y=f(-x)\) to reflect in the \(y\)-axis (negate \(x\): \((x,y)\mapsto(-x,y)\)). The minus outside acts on \(y\); the minus inside acts on \(x\).

Does a dilation change the x-intercepts?

A dilation from the \(x\)-axis with \(a>0\) leaves them fixed, since \(a\times 0=0\). A dilation from the \(y\)-axis divides them by \(n\), so \(y=3f(x)\) keeps the \(x\)-intercepts but \(y=f(3x)\) moves them to one third of their \(x\)-values.

What happens to asymptotes under a dilation or reflection?

A horizontal asymptote \(y=k\) becomes \(y=ak\) under \(y=a\,f(x)\) and changes sign under a reflection in the \(x\)-axis. A vertical asymptote \(x=c\) is divided by \(n\) under \(y=f(nx)\) and changes sign under a reflection in the \(y\)-axis; an asymptote at \(y=0\) or \(x=0\) stays put.

How do you find the image of a point?

Apply the mapping rule: \((x,y)\mapsto(x,ay)\) for \(y=a\,f(x)\), \(\left(\dfrac{x}{n},y\right)\) for \(y=f(nx)\), \((x,-y)\) for \(y=-f(x)\), and \((-x,y)\) for \(y=f(-x)\). Transform the key points, intercepts and asymptotes, then sketch.

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