This note covers graph transformations that go beyond translations and stretches: the modulus graphs \(y=|f(x)|\) and \(y=f(|x|)\), the reciprocal graph \(y=\dfrac1{f(x)}\), the square graph \(y=[f(x)]^2\), and the composite horizontal transformation \(y=f(ax+b)\). They let you sketch a new graph from the original one without any calculation.
Key ideas
The graph of \(y=|f(x)|\). The modulus makes every output non-negative. Keep the parts of the graph of \(y=f(x)\) that are on or above the \(x\)-axis. Reflect the parts below the \(x\)-axis in the \(x\)-axis, so they flip upward. The \(x\)-intercepts stay where they are, and the graph never goes below the axis. Where the reflection meets the axis, there is usually a sharp corner.
The graph of \(y=f(|x|)\). The modulus is now on the input. Keep only the part of the graph where \(x\ge0\), delete the part where \(x\lt0\), and replace it with a reflection of the right-hand part in the \(y\)-axis. The result is symmetric about the \(y\)-axis, and the \(y\)-intercept is unchanged.
The graph of \(y=\dfrac1{f(x)}\). Each output is replaced by its reciprocal. Use these facts:
- Where \(f(x)=0\), the reciprocal is undefined, so there is a vertical asymptote. Zeros of \(f\) become vertical asymptotes.
- The sign is kept: where \(f\) is positive, so is \(1/f\), and the same for negative.
- Where \(f(x)=1\) or \(f(x)=-1\), the reciprocal has the same value, so those points stay on the graph.
- When \(f(x)\) is large, \(1/f(x)\) is close to zero. So \(y=0\) is a horizontal asymptote if \(f\) becomes large.
- Increasing sections of \(f\) become decreasing sections of \(1/f\). A maximum of \(f\) gives a minimum of \(1/f\) (when \(f\) is positive there), and a minimum of \(f\) gives a maximum of \(1/f\) (when \(f\) is negative there).
- Where \(f\) has a vertical asymptote, \(1/f\) has a zero.
The graph of \(y=[f(x)]^2\). Squaring makes every output non-negative. Zeros of \(f\) stay as zeros, where the graph touches the \(x\)-axis. Points where \(f=1\) or \(f=-1\) both go to height 1. Values with \(|f|\gt1\) get larger, and values with \(|f|\lt1\) get smaller. Turning points of \(f\) stay at the same \(x\)-values. Parts of the graph that were below the axis end up above it.
The graph of \(y=f(ax+b)\). Factorise to get \(f\big(a(x+\tfrac ba)\big)\). The transformation is a horizontal stretch with scale factor \(\dfrac1a\) (with a reflection in the \(y\)-axis if \(a\lt0\)), followed by a translation of \(-\dfrac ba\) in the \(x\)-direction. Using the point rule, if \((p,q)\) is on \(y=f(x)\), then the image has \(ax+b=p\), so \(x=\dfrac{p-b}{a}\), and \(y=q\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(|a|=a\) if \(a\ge0\); \(|a|=-a\) if \(a\lt0\) | Modulus (absolute value). It is never negative. |
| \(y=|f(x)|\) | Reflect the parts of \(y=f(x)\) below the \(x\)-axis in the \(x\)-axis. |
| \(y=f(|x|)\) | Keep the \(x\ge0\) part, then reflect it in the \(y\)-axis. |
| \(y=\dfrac1{f(x)}\) | Zeros of \(f\) give vertical asymptotes. Points with \(f=\pm1\) are invariant. Large \(f\) gives \(y\) near 0. |
| \(y=[f(x)]^2\) | Output squared. Non-negative. Zeros stay as touching points. Points with \(f=\pm1\) go to 1. |
| \(y=f(ax+b)\) | Horizontal stretch (factor \(\frac1a\)), then translation \(-\frac ba\) in the \(x\)-direction. Point \((p,q)\to\left(\frac{p-b}a,\,q\right)\). |
Worked examples
Example 1 The function \(f(x)=x^2-x-6\) has its minimum at \((0.5,-6.25)\). For the graphs of \(y=|f(x)|\) and \(y=f(|x|)\), find the intercepts and the turning points.
Factorise: \(f(x)=(x-3)(x+2)\). So the \(x\)-intercepts of \(f\) are \(-2\) and \(3\), and \(f(0)=-6\).
For \(y=|f(x)|\): the part of the parabola between \(x=-2\) and \(x=3\) is below the axis, so it is reflected upward. The \(x\)-intercepts stay at \((-2,0)\) and \((3,0)\). The \(y\)-intercept becomes \(|-6|=6\), so \((0,6)\). The minimum point \((0.5,-6.25)\) is reflected to \((0.5,6.25)\), which is now a local maximum.
For \(y=f(|x|)=x^2-|x|-6\): keep the right half, where \(x\ge0\). The \(y\)-intercept is unchanged at \((0,-6)\). On the right, the \(x\)-intercept is \((3,0)\), and the minimum is \((0.5,-6.25)\). Reflect in the \(y\)-axis to get the left half: the \(x\)-intercepts are \((\pm3,0)\), and the minima are \((\pm0.5,-6.25)\). At \(x=0\), the graph has a local maximum at \((0,-6)\), with a sharp point where the two halves meet.
Answer: \(|f|\): intercepts \((-2,0)\), \((3,0)\), \((0,6)\); local maximum \((0.5,6.25)\). \(f(|x|)\): intercepts \((\pm3,0)\), \((0,-6)\); minima \((\pm0.5,-6.25)\); local maximum \((0,-6)\).
Example 2 Let \(f(x)=x^2-2x-3\). Sketch the features of \(y=\dfrac1{f(x)}\): asymptotes, intercepts and turning point.
Factorise: \(f(x)=(x-3)(x+1)\). The zeros are \(x=3\) and \(x=-1\), so \(y=\dfrac1{f(x)}\) has vertical asymptotes \(x=3\) and \(x=-1\).
As \(x\to\pm\infty\), \(f(x)\to\infty\), so \(\dfrac1{f(x)}\to0\). The horizontal asymptote is \(y=0\). Since \(f\) is never infinite, \(1/f\) has no zero, so there is no \(x\)-intercept.
The \(y\)-intercept is \(\dfrac1{f(0)}=\dfrac{1}{-3}=-\dfrac13\).
The minimum of \(f\) is at \(x=1\), where \(f(1)=1-2-3=-4\). It is negative, so the reciprocal has a local maximum there, at \(\left(1,-\dfrac14\right)\).
Signs: \(f\) is positive for \(x\lt-1\) and \(x\gt3\), so \(1/f\) is positive there, and falls towards 0 as \(|x|\) grows. Between the asymptotes \(f\lt0\), so \(1/f\) is negative and goes down to \(-\infty\) at both asymptotes, with its highest point \(-\tfrac14\) at \(x=1\).
Answer: VA \(x=-1\) and \(x=3\); HA \(y=0\); \(y\)-intercept \(\left(0,-\tfrac13\right)\); local maximum \(\left(1,-\tfrac14\right)\); no \(x\)-intercept
Example 3 The graph of \(y=f(x)\) has a local maximum at \((1,4)\), a local minimum at \((5,-2)\), and passes through \((3,0)\). Find the image of each of these three points on (a) \(y=[f(x)]^2\) and (b) \(y=f(2x+6)\).
(a) Square each \(y\)-coordinate and keep \(x\). \((1,4)\to(1,16)\). \((3,0)\to(3,0)\). \((5,-2)\to(5,4)\).
The point \((3,0)\) is where the graph touches the axis from above. The point \((5,-2)\) becomes a local maximum, because the values of \(f\) near it are between \(-2\) and \(0\), and squaring them makes the graph fall away from 4 on either side.
(b) Solve \(2x+6=p\) for each \(x\)-coordinate \(p\), so \(x=\dfrac{p-6}{2}\), and keep \(y\).
\(p=1\): \(x=\dfrac{-5}{2}=-2.5\). Image \((-2.5,4)\).
\(p=3\): \(x=\dfrac{-3}{2}=-1.5\). Image \((-1.5,0)\).
\(p=5\): \(x=\dfrac{-1}{2}=-0.5\). Image \((-0.5,-2)\).
Check with the transformation: \(f(2(x+3))\) is a horizontal stretch of scale factor \(\tfrac12\) followed by a translation 3 units left. For \(p=1\): \(1\to0.5\to-2.5\). This matches.
Answer: (a) \((1,16)\), \((3,0)\), \((5,4)\). (b) \((-2.5,4)\), \((-1.5,0)\), \((-0.5,-2)\).
Common errors
- Confusing \(|f(x)|\) with \(f(|x|)\). The first reflects the part below the \(x\)-axis. The second reflects the right-hand part across the \(y\)-axis.
- Placing the vertical asymptotes of \(1/f\) at the wrong places. They are at the \(x\)-values where \(f(x)=0\), not at the \(x\)-values where \(f\) has turning points.
- Forgetting that the turning points of \(1/f\) are at the same \(x\)-values as those of \(f\), but have reciprocal \(y\)-values (\(4\) becomes \(\tfrac14\), \(-4\) becomes \(-\tfrac14\)).
- For \(f(2x+6)\), stretching by \(\tfrac12\) and then translating 6 left. That gives \(f(2(x+6))=f(2x+12)\). The correct version is a stretch by \(\tfrac12\) then a translation 3 left, or a translation 6 left then the stretch.
- Showing \([f(x)]^2\) with parts below the \(x\)-axis. The square is never negative.
Exam tip: When sketching \(y=\dfrac1{f(x)}\), first mark the zeros of \(f\) as dashed vertical lines and the points where \(f=\pm1\). Then draw each branch between them.
Check yourself
- Write \(|f(x)|\) in the case \(f(x)=2x-4\) without the modulus symbol, and state the vertex.
Answer
\(|2x-4|=2x-4\) for \(x\ge2\), and \(-(2x-4)=4-2x\) for \(x\lt2\). The graph is a V-shape with its vertex at \((2,0)\). - For \(f(x)=x-2\), find the \(x\)-intercepts and \(y\)-intercept of \(y=f(|x|)\).
Answer
\(f(|x|)=|x|-2\). When \(x=0\): \(y=-2\), so the \(y\)-intercept is \((0,-2)\). When \(y=0\): \(|x|=2\), so \(x=\pm2\), giving intercepts \((-2,0)\) and \((2,0)\). - State the asymptotes and the \(y\)-intercept of \(y=\dfrac1{f(x)}\) for \(f(x)=x^2-9\).
Answer
\(x^2-9=(x-3)(x+3)\), so the vertical asymptotes are \(x=3\) and \(x=-3\). As \(|x|\) grows, \(1/f\to0\), so \(y=0\) is a horizontal asymptote. The \(y\)-intercept is \(\dfrac{1}{f(0)}=\dfrac{1}{-9}\), the point \(\left(0,-\tfrac19\right)\). - Describe the transformation from \(y=f(x)\) to \(y=f(2x+4)\).
Answer
\(f(2x+4)=f\big(2(x+2)\big)\). It is a horizontal stretch with scale factor \(\tfrac12\) (parallel to the \(x\)-axis), followed by a translation of \(\begin{pmatrix}-2\\0\end{pmatrix}\).