International Baccalaureate · DP 1: Analysis and Approaches HL

Transformations of Functions

⌚ ~130 min ▢ 14 blocks ✓ 63 exercises ➣ Prerequisites : Functions, Functions

I Translations, Stretches and Reflections

I.1 Translations

Recappreviously studied — click to reread

Function transformations take a basic “parent” function and modify it to create a new, related function. A sequence of transformations can translate, stretch or reflect the graph of the parent function.
Understanding these transformations makes it possible to predict the graph of a more complicated function from a simpler one, and to model real-world phenomena by adjusting a basic function to fit observed data. The transformations are vertical and horizontal translations, stretches (a stretch with a scale factor between $0$ and $1$ is also called a compression) and reflections.

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Definition — Vertical Translation by Adding a Constant

A vertical translation shifts the graph of a function up or down. The transformation is defined by:$$g(x) = f(x) + k$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x, y+k)$ on the graph of $g$.

If $k>0$, the graph is translated $\boldsymbol{k}$ units up.
If $k<0$, the graph is translated $\boldsymbol{|k|}$ units down.

Definition — Horizontal Translation

A horizontal translation shifts the graph of a function left or right. The transformation is defined by:$$g(x) = f(x-h)$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x+h, y)$ on the graph of $g$.

If $h>0$, the graph is translated $\boldsymbol{h}$ units to the right.
If $h<0$, the graph is translated $\boldsymbol{|h|}$ units to the left.

Example

From the graph of $f(x) = x^2$, plot $g(x) = f(x) - 3$ and $h(x) = f(x + 1)$.

$g(x) = f(x) - 3$: every output is lowered by $3$ — a vertical translation $3$ units down. The vertex $(0, 0)$ moves to $(0, -3)$.
$h(x) = f(x + 1)$: the input $x = -1$ now plays the role $0$ used to play — a horizontal translation $1$ unit to the left. The vertex $(0, 0)$ moves to $(-1, 0)$.

Translation by a Vector

The vertical and the horizontal translations combine into one. The translation by the vector $\begin{pmatrix} h \\ k \end{pmatrix}$ moves each point $(x, y)$ of the graph of $f$ to the point $(X, Y) = (x + h, y + k)$. Since $x = X - h$ and $Y = f(x) + k$, the image points satisfy $Y = f(X - h) + k$; writing $(x, y)$ again for the coordinates, the image of the graph of $y = f(x)$ is the graph of $$y = f(x - h) + k.$$ For example, the translation by the vector $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ maps the graph of $y = x^2$ onto the graph of $y = (x - 3)^2 - 2$, and its vertex $(0, 0)$ to $(3, -2)$.

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Skills to practice

Translating Graphs

I.2 Stretches

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Definition — Vertical Stretch

A vertical stretch stretches or compresses the graph of a function vertically. It is defined, for a number $a \neq 0$, by:$$g(x) = a \cdot f(x)$$ This transformation maps a point $(x, y)$ to $(x, ay)$.

If $|a|>1$, the graph is stretched vertically by a factor of $|a|$.
If $0 < |a| < 1$, the graph is compressed vertically by a factor of $|a|$.
If $a < 0$, the stretch or compression is followed by a reflection in the $x$-axis ($a = -1$ gives the reflection alone).

Definition — Horizontal Stretch

A horizontal stretch stretches or compresses the graph of a function horizontally. It is defined, for a number $b \neq 0$, by:$$g(x) = f(bx)$$ This transformation maps a point $(x, y)$ to a new point $(\frac{x}{b}, y)$.

If $|b|>1$, the graph is compressed horizontally by a factor of $\frac{1}{|b|}$.
If $0 < |b| < 1$, the graph is stretched horizontally by a factor of $\frac{1}{|b|}$.
If $b < 0$, the stretch or compression is followed by a reflection in the $y$-axis ($b = -1$ gives the reflection alone).

Example

From the graph of $f(x) = x^2$, plot $a(x) = 2f(x)$ and $b(x) = f(2x)$.

$a(x) = 2f(x) = 2x^2$: every output is doubled — a vertical stretch with scale factor $2$. The point $(1, 1)$ moves to $(1, 2)$.
$b(x) = f(2x) = 4x^2$: the input is consumed twice as fast — a horizontal stretch with scale factor $\frac12$, a compression: every $x$-coordinate is halved. The point $(1, 1)$ moves to $\left(\frac12, 1\right)$.

The two results differ: doubling the output is not the same as doubling the input.

Stretches and Scale Factors

The number by which a stretch multiplies the distances to the axis that stays fixed is its scale factor. For $a > 0$, the graph of $y = a f(x)$ is the image of the graph of $y = f(x)$ under the vertical stretch with scale factor $a$: each point $(x, y)$ goes to $(x, ay)$. For $b > 0$, the graph of $y = f(bx)$ is its image under the horizontal stretch with scale factor $\dfrac{1}{b}$: each point $(x, y)$ goes to $\left(\dfrac{x}{b}, y\right)$. A scale factor greater than $1$ moves the points away from the axis that stays fixed (the $x$-axis for a vertical stretch, the $y$-axis for a horizontal one), a scale factor between $0$ and $1$ brings them closer to it: the graph of $y = f(2x)$ is the image of the graph of $f$ under the horizontal stretch with scale factor $\dfrac{1}{2}$, a compression.

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Skills to practice

Stretching Graphs

I.3 Reflections

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Definition — Reflection in the x-axis

A reflection in the x-axis flips the graph of a function vertically. It is a special case of vertical stretch where $a=-1$. It is defined by:$$g(x) = -f(x)$$ This transformation maps a point $(x, y)$ to $(x, -y)$.

Definition — Reflection in the y-axis

A reflection in the y-axis flips the graph of a function horizontally. It is a special case of horizontal stretch where $b=-1$. It is defined by:$$g(x) = f(-x)$$ This transformation maps a point $(x, y)$ to $(-x, y)$.

Example

From the graph of $f(x) = \sqrt{x}$, plot $r(x) = -f(x)$ and $s(x) = f(-x)$.

$r(x) = -\sqrt{x}$: every output changes sign — a reflection in the $x$-axis. The curve flips below the axis; the point $(4, 2)$ moves to $(4, -2)$.
$s(x) = \sqrt{-x}$: every input changes sign — a reflection in the $y$-axis. The curve flips to the left half-plane (defined for $x \le 0$); the point $(4, 2)$ moves to $(-4, 2)$.

Skills to practice

Reflecting Graphs

II Composite Transformations

II.1 Combining Transformations

Recappreviously studied — click to reread

When several transformations are applied to a function, the order can matter between two horizontal transformations, or between two vertical ones (a stretch and a translation, for example); a horizontal transformation and a vertical one can be applied in either order. The standard form for a transformed function is:$$ g(x) = a \cdot f(b(x-c)) + d $$ To graph this function from the parent function $f(x)$, apply the transformations in the following order:

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Method — Order of Transformations

Horizontal Transformations (inside the brackets):
- Apply the horizontal stretch with scale factor $\frac{1}{|b|}$ ($b \neq 0$), followed by a reflection in the $y$-axis if $b < 0$.
- Apply the horizontal translation (shift) by $c$ units.
Vertical Transformations (outside the brackets):
- Apply the vertical stretch with scale factor $|a|$, followed by a reflection in the $x$-axis if $a < 0$.
- Apply the vertical translation (shift) by $d$ units.

Note: Always take out the coefficient $b$ as a factor of the term inside the function to correctly identify the horizontal shift $c$. For example, transform $f(2x-6)$ as $f(2(x-3))$. This shows the horizontal stretch with scale factor $\frac{1}{2}$ (a compression) followed by a translation of 3 units to the right, not 6.

Example

Describe the sequence of transformations that maps the graph of $f(x)=\sqrt{x}$ onto the graph of $g(x) = 3\sqrt{-x+2} - 4$.

Answer

First, we rewrite the function $g(x)$ in the standard form $a \cdot f(b(x-c)) + d$:$$ g(x) = 3\sqrt{-(x-2)} - 4 $$ Comparing this to $f(x)=\sqrt{x}$, we have $a=3$, $b=-1$, $c=2$, and $d=-4$. The transformations are applied as follows:

Horizontal Transformations on $f(x)=\sqrt{x}$:
- Reflection: Since $b=-1$, reflect the graph in the y-axis ($y = \sqrt{-x}$).
- Translation: Since $c=2$, translate the graph 2 units to the right ($y = \sqrt{-(x-2)}$).
Vertical Transformations on the result:
- Stretch: Since $a=3$, stretch the graph vertically with scale factor 3 ($y = 3\sqrt{-(x-2)}$).
- Translation: Since $d=-4$, translate the graph 4 units down ($y = 3\sqrt{-(x-2)} - 4$).

Example

The graph of $y = f(x)$ is reflected in the $y$-axis, then stretched horizontally with scale factor $2$. Find an equation of the image, and check it with a point $(a, b)$ of the graph of $f$.

Answer

Each transformation acts on the equation obtained at the previous step: a horizontal transformation replaces $x$ in the current expression, a vertical transformation acts on the whole current expression.

Reflection in the $y$-axis: $x$ is replaced by $-x$: $y = f(-x)$.
Horizontal stretch with scale factor $2$: $x$ is replaced by $\dfrac{x}{2}$ in the current expression $f(-x)$: $y = f\left(-\dfrac{x}{2}\right)$.

The image is the graph of $y = f\left(-\dfrac{x}{2}\right)$.
Check with a point. The reflection sends a point $(a, b)$ of the graph of $f$ to $(-a, b)$, and the stretch doubles the $x$-coordinate: $(-a, b)$ goes to $(-2a, b)$. At $x = -2a$, the expression inside $f$ is $$\begin{aligned}[t] -\frac{x}{2} &= -\frac{-2a}{2} && (x = -2a) \\ &= a, \end{aligned}$$ so $f\left(-\dfrac{x}{2}\right) = f(a) = b$, since the point $(a, b)$ is on the graph of $f$: the point $(-2a, b)$ is indeed on the image. Replacing $x$ by $-2x$ instead, that is writing $f(-2x)$, would give the horizontal stretch with scale factor $\dfrac{1}{2}$: the check with a point detects this error, since $f\big(-2 \times (-2a)\big) = f(4a)$ is not equal to $b$ in general.

Skills to practice

Sketching Combined Transformations
Finding Equations after a Sequence of Transformations

II.2 Rational Functions as Images of $y \equal \frac{1}{x}$

A function of the form $y = \dfrac{ax + b}{cx + d}$, with $c \neq 0$ and $ad - bc \neq 0$, can be rewritten as $y = p + \dfrac{q}{x - h}$ by writing the numerator around the denominator. Its graph is then the image of the graph of $y = \dfrac{1}{x}$ under a sequence of translations, stretches and reflections, and its asymptotes are the images of the two axes.

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Example

Describe a sequence of transformations that maps the graph of $y = \dfrac{1}{x}$ onto the graph of $y = \dfrac{2x + 1}{x - 1}$, and deduce the asymptotes of the second graph.

Answer

Write the numerator around the denominator: $2x + 1 = 2(x - 1) + 3$. With $f(x) = \dfrac{1}{x}$, for $x \neq 1$: $$\begin{aligned}[t] \frac{2x + 1}{x - 1} &= \frac{2(x - 1) + 3}{x - 1} \\ &= 2 + \frac{3}{x - 1} && \text{(split the fraction)} \\ &= 3f(x - 1) + 2. \end{aligned}$$ This is the standard form $a \cdot f(b(x - c)) + d$ with $a = 3$, $b = 1$, $c = 1$ and $d = 2$. In the order of transformations:

translate $1$ unit to the right: $y = \dfrac{1}{x - 1}$;
stretch vertically with scale factor $3$: $y = \dfrac{3}{x - 1}$;
translate $2$ units up: $y = \dfrac{3}{x - 1} + 2$.

The asymptotes $x = 0$ and $y = 0$ of $y = \dfrac{1}{x}$ follow the graph: the translation $1$ unit to the right moves them to $x = 1$ and $y = 0$; the vertical stretch leaves these two lines in place; the translation $2$ units up moves $y = 0$ to $y = 2$. The graph of $y = \dfrac{2x + 1}{x - 1}$ has the asymptotes $x = 1$ and $y = 2$. The point $(1, 1)$ of $y = \dfrac{1}{x}$ goes to $(2, 1)$, then $(2, 3)$, then $(2, 5)$; check: $\dfrac{2 \times 2 + 1}{2 - 1} = 5$.

Skills to practice

Relating Rational Functions to $y \equal \frac{1}{x}$

III The Graph of $y \equal \frac{1}{f(x)}$

III.1 Reciprocal Graphs

Discover

Let $f(x) = x^2 - 4$. Whenever $f(x) \neq 0$, the number $f(x)$ has a reciprocal $\dfrac{1}{f(x)}$: how does the graph of $y = \dfrac{1}{f(x)}$ compare with the graph of $f$?

Compute $f(x)$ and $\dfrac{1}{f(x)}$ for $x = 0$, $1$, $3$ and $4$. Why are the values the same for $x = -1$, $-3$ and $-4$?
Why can $\dfrac{1}{f(x)}$ not be computed for $x = 2$ and for $x = -2$? What happens to $\dfrac{1}{f(x)}$ when $x$ gets closer and closer to $2$?
What happens to $\dfrac{1}{f(x)}$ when $x$ becomes large?
For which values of $x$ are the two numbers $f(x)$ and $\dfrac{1}{f(x)}$ equal?

Answer

The values:

$x$ $0$ $1$ $3$ $4$

$f(x) = x^2 - 4$ $-4$ $-3$ $5$ $12$

$\dfrac{1}{f(x)}$ $-\dfrac{1}{4}$ $-\dfrac{1}{3}$ $\dfrac{1}{5}$ $\dfrac{1}{12}$

Since $(-x)^2 = x^2$, $f(-x) = f(x)$: the values at $-1$, $-3$ and $-4$ are those at $1$, $3$ and $4$. Each reciprocal has the same sign as $f(x)$, and the larger $|f(x)|$ is, the smaller $\left|\dfrac{1}{f(x)}\right|$ is.
$f(2) = f(-2) = 0$, and $0$ has no reciprocal. Close to $2$, $f(x)$ is close to $0$: $f(2.1) = 0.41$, $f(2.01) = 0.0401$ and $f(1.99) = -0.0399$, so $\dfrac{1}{f(2.1)} \approx 2.44$, $\dfrac{1}{f(2.01)} \approx 24.9$ and $\dfrac{1}{f(1.99)} \approx -25.1$. The absolute values $\left|\dfrac{1}{f(x)}\right|$ become as large as desired: the line $x = 2$ is a vertical asymptote of the graph of $y = \dfrac{1}{f(x)}$, and so is the line $x = -2$.
When $x$ becomes large, $f(x) = x^2 - 4$ becomes as large as desired, so $\dfrac{1}{f(x)}$ gets as close to $0$ as desired: the $x$-axis is a horizontal asymptote.
For $f(x) \neq 0$: $$\begin{aligned}[t] f(x) = \frac{1}{f(x)} &\iff f(x)^2 = 1 && \text{(multiply by } f(x) \neq 0\text{)} \\ &\iff f(x) = 1 \ \text{ or } \ f(x) = -1 \\ &\iff x^2 = 5 \ \text{ or } \ x^2 = 3 && (f(x) = x^2 - 4) \\ &\iff x = \pm\sqrt{5} \ \text{ or } \ x = \pm\sqrt{3}. \end{aligned}$$ The two graphs meet at four points, on the lines $y = 1$ and $y = -1$.

The graph of $y = \dfrac{1}{f(x)}$ has three branches, separated by the vertical asymptotes $x = -2$ and $x = 2$; the minimum point $(0, -4)$ of $f$ has become the highest point $\left(0, -\dfrac{1}{4}\right)$ of the middle branch. The Propositions “Sign, Invariant Points and Variations” and “Asymptotes of the Graph of $y = \dfrac{1}{f(x)}$” turn each of these observations into a rule.

Definition — The Reciprocal of a Function

Let $f$ be a function. The reciprocal of $f$ is the function $\dfrac{1}{f}$ given by $$\frac{1}{f}(x) = \frac{1}{f(x)}.$$ It is defined for every $x$ of the domain of $f$ such that $f(x) \neq 0$. Its graph is the graph of the equation $y = \dfrac{1}{f(x)}$. A point lying on both the graph of $f$ and the graph of $y = \dfrac{1}{f(x)}$ is an invariant point.
The reciprocal $\dfrac{1}{f}$ must not be confused with the inverse function $f^{-1}$, which undoes $f$: for $f(x) = 2x$, $\dfrac{1}{f(x)} = \dfrac{1}{2x}$, whereas $f^{-1}(x) = \dfrac{x}{2}$.

Proposition — Sign, Invariant Points and Variations

Let $f$ be a function.

Sign: wherever $f(x) \neq 0$, the number $\dfrac{1}{f(x)}$ has the same sign as $f(x)$. The graphs of $y = f(x)$ and $y = \dfrac{1}{f(x)}$ lie on the same side of the $x$-axis.
Invariant points: a point lies on both graphs exactly when it is a point of the graph of $f$ at which $f(x) = 1$ or $f(x) = -1$. The two graphs meet only on the lines $y = 1$ and $y = -1$.
Variations: on an interval where $f(x)$ is never $0$ and keeps the same sign, if $f$ is increasing then $\dfrac{1}{f}$ is decreasing, and if $f$ is decreasing then $\dfrac{1}{f}$ is increasing.
Maximum and minimum: on such an interval, if $f(a)$ is the maximum of $f$, then $\dfrac{1}{f(a)}$ is the minimum of $\dfrac{1}{f}$, and if $f(a)$ is the minimum of $f$, then $\dfrac{1}{f(a)}$ is the maximum of $\dfrac{1}{f}$. So, when the graph of $f$ has no break near $a$ (then $f(x)$ keeps the sign of $m$ for $x$ close to $a$), a local maximum point $(a, m)$ of the graph of $f$, with $m \neq 0$, becomes a local minimum point $\left(a, \dfrac{1}{m}\right)$ of the graph of $y = \dfrac{1}{f(x)}$, and a local minimum point becomes a local maximum point.

Proof

Sign. $f(x) \times \dfrac{1}{f(x)} = 1 > 0$: the two numbers are nonzero and have the same sign.
Invariant points. The point with $x$-coordinate $x$ lies on both graphs when $f(x) = \dfrac{1}{f(x)}$, with $f(x) \neq 0$: $$\begin{aligned}[t] f(x) = \frac{1}{f(x)} &\iff f(x)^2 = 1 && \text{(multiply by } f(x) \neq 0\text{)} \\ &\iff f(x) = 1 \ \text{ or } \ f(x) = -1. \end{aligned}$$
Variations. Let $u \leq v$ be two numbers of an interval $I$ on which $f(x)$ is never $0$ and keeps the same sign. Then $f(u) f(v) > 0$, and $$\begin{aligned}[t] \frac{1}{f(u)} - \frac{1}{f(v)} &= \frac{f(v) - f(u)}{f(u) f(v)}. && \text{(common denominator)} \end{aligned}$$ If $f$ is increasing, $f(v) - f(u) \geq 0$, so $\dfrac{1}{f(u)} \geq \dfrac{1}{f(v)}$: $\dfrac{1}{f}$ is decreasing on $I$. If $f$ is decreasing, $f(v) - f(u) \leq 0$ and $\dfrac{1}{f}$ is increasing on $I$.
Maximum and minimum. If $f(a)$ is the maximum of $f$ on $I$, then $f(x) \leq f(a)$ for every $x$ of $I$, and $f(x) f(a) > 0$, so $$\frac{1}{f(x)} - \frac{1}{f(a)} = \frac{f(a) - f(x)}{f(x) f(a)} \geq 0:$$ $\dfrac{1}{f(a)}$ is the minimum of $\dfrac{1}{f}$ on $I$. The case of a minimum is the same with the inequalities reversed. When the graph of $f$ has no break at $a$ and $f(a) = m \neq 0$, $f(x)$ keeps the sign of $m$ on an interval around $a$, on which this applies: a local maximum point of the graph of $f$ becomes a local minimum point, and conversely.

Proposition — Asymptotes of the Graph of $y \equal \frac{1}{f(x)}$

Let $f$ be a function whose graph has no break, except at its vertical asymptotes.

Zeros become vertical asymptotes: if $f(a) = 0$, the graph of $f$ has no break at $a$, and $f(x) \neq 0$ for the other values of $x$ close to $a$, the line $x = a$ is a vertical asymptote of the graph of $y = \dfrac{1}{f(x)}$.
Large values become values close to $0$: where $|f(x)|$ becomes as large as desired, $\dfrac{1}{f(x)}$ gets as close to $0$ as desired. So if $|f(x)|$ becomes as large as desired when $x$ becomes large, the $x$-axis is a horizontal asymptote of the graph of $y = \dfrac{1}{f(x)}$; and if the line $x = p$ is a vertical asymptote of the graph of $f$, the graph of $y = \dfrac{1}{f(x)}$ comes as close as desired to the point $(p, 0)$, which does not belong to it, since a reciprocal $\dfrac{1}{f(x)}$ is never $0$: this point is drawn as an open circle.
Horizontal asymptotes: if the line $y = q$, with $q \neq 0$, is a horizontal asymptote of the graph of $f$, the line $y = \dfrac{1}{q}$ is a horizontal asymptote of the graph of $y = \dfrac{1}{f(x)}$.

Proof

Zeros. Since the graph of $f$ has no break at $a$ and $f(a) = 0$, $f(x)$ gets as close to $0$ as desired when $x$ gets close to $a$, without being $0$ for $x \neq a$. Then $\left|\dfrac{1}{f(x)}\right| = \dfrac{1}{|f(x)|}$ becomes as large as desired: by the definition of a vertical asymptote, the line $x = a$ is one.
Large values. If $|f(x)|$ is larger than a number $M > 0$, then $\left|\dfrac{1}{f(x)}\right| = \dfrac{1}{|f(x)|}$ is smaller than $\dfrac{1}{M}$, which is as small as desired when $M$ is large enough.
Horizontal asymptotes. Suppose $f(x)$ gets as close to $q \neq 0$ as desired when $x$ becomes large. Once $f(x)$ is within $\dfrac{|q|}{2}$ of $q$, $|f(x)| \geq \dfrac{|q|}{2}$, and $$\left|\frac{1}{f(x)} - \frac{1}{q}\right| = \frac{|q - f(x)|}{|q| \, |f(x)|} \leq \frac{2}{q^2}\,|q - f(x)|,$$ which gets as close to $0$ as desired: $\dfrac{1}{f(x)}$ gets as close to $\dfrac{1}{q}$ as desired, and the line $y = \dfrac{1}{q}$ is a horizontal asymptote.

Method — Sketching the Graph of $y \equal \frac{1}{f(x)}$ from the Graph of $y \equal f(x)$

Draw the vertical asymptotes: the lines $x = a$ at the zeros $a$ of $f$ (each zero $a$ such that $f(x) \neq 0$ for the other values of $x$ close to $a$).
Mark the invariant points, where the graph of $f$ meets the lines $y = 1$ and $y = -1$.
Turn each turning point $(a, m)$ of $f$, with $m \neq 0$ and the graph of $f$ without a break near $a$, into $\left(a, \dfrac{1}{m}\right)$, a maximum becoming a minimum and a minimum a maximum; mark the $y$-intercept $\left(0, \dfrac{1}{f(0)}\right)$ when $f(0) \neq 0$.
Where $|f(x)|$ becomes large, draw $\dfrac{1}{f(x)}$ close to $0$: the $x$-axis as a horizontal asymptote, or an open circle at $(p, 0)$ for a vertical asymptote $x = p$ of $f$. A horizontal asymptote $y = q \neq 0$ of $f$ gives the horizontal asymptote $y = \dfrac{1}{q}$.
Join the points on each interval between the asymptotes, leaving out the open circles, keeping the sign of $f$ and reversing its variations: where $f$ increases, $\dfrac{1}{f}$ decreases, and where $f$ decreases, $\dfrac{1}{f}$ increases.
The range of $\dfrac{1}{f}$ is made of the numbers $\dfrac{1}{y}$, for the nonzero values $y$ taken by $f$.

Example

Sketch the graph of $y = \dfrac{1}{2x - 4}$ from the graph of $f(x) = 2x - 4$.

Answer

Vertical asymptote. $2x - 4 = 0$ for $x = 2$: the line $x = 2$.
Invariant points. $2x - 4 = 1$ for $x = \dfrac{5}{2}$, and $2x - 4 = -1$ for $x = \dfrac{3}{2}$: the points $\left(\dfrac{5}{2}, 1\right)$ and $\left(\dfrac{3}{2}, -1\right)$.
Intercept. $\dfrac{1}{f(0)} = \dfrac{1}{-4} = -\dfrac{1}{4}$: the point $\left(0, -\dfrac{1}{4}\right)$. A line that is not horizontal has no turning point.
Large values. $|2x - 4|$ becomes as large as desired when $x$ becomes large: the $x$-axis is a horizontal asymptote.
Sign and variations. $f$ is increasing, negative for $x < 2$ and positive for $x > 2$: $\dfrac{1}{f}$ is decreasing on each of the two intervals. The left branch lies below the $x$-axis and falls along $x = 2$; the right branch lies above it and comes down from the top along $x = 2$.

Example

The graph of a quadratic function $f$ is shown: it cuts the $x$-axis at $-1$ and $3$, and its vertex is $(1, -4)$. Sketch the graph of $y = \dfrac{1}{f(x)}$.

Answer

Vertical asymptotes. The zeros of $f$ are $-1$ and $3$: the lines $x = -1$ and $x = 3$.
Turning point. The vertex $(1, -4)$ is the minimum point of $f$, with $-4 \neq 0$: it becomes the local maximum point $\left(1, -\dfrac{1}{4}\right)$.
Invariant points. The parabola meets each of the lines $y = 1$ and $y = -1$ at two points, which lie on both graphs. With the formula $f(x) = (x + 1)(x - 3)$, found from the intercepts and the vertex (the coefficient $a$ of $f(x) = a(x + 1)(x - 3)$ satisfies $f(1) = -4a = -4$, so $a = 1$), their $x$-coordinates are $1 \pm \sqrt{5}$ and $1 \pm \sqrt{3}$. The $y$-intercept $f(0) = -3$ gives the point $\left(0, -\dfrac{1}{3}\right)$.
Large values. $f(x)$ becomes as large as desired when $x$ becomes large: the $x$-axis is a horizontal asymptote.
Sign and variations. $f(x) > 0$ for $x < -1$ and for $x > 3$, $f(x) < 0$ between $-1$ and $3$. The outer branches lie above the $x$-axis and climb along the asymptotes; the middle branch lies below it, rises from the bottom along $x = -1$ to its highest point $\left(1, -\dfrac{1}{4}\right)$, then falls along $x = 3$.

Example

The graph of $f(x) = \dfrac{2x - 2}{x + 1}$ is shown, with its asymptotes $x = -1$ and $y = 2$; it passes through $(1, 0)$ and $(0, -2)$. Give the domain of $\dfrac{1}{f}$ and sketch the graph of $y = \dfrac{1}{f(x)}$.

Answer

Domain. $\dfrac{1}{f(x)}$ needs $f(x)$ to be defined, so $x \neq -1$, and nonzero, so $x \neq 1$: the domain is $\mathbb{R} \setminus \{-1, 1\}$.
Asymptotes. The zero $1$ of $f$ gives the vertical asymptote $x = 1$. The horizontal asymptote $y = 2$ of $f$ gives the horizontal asymptote $y = \dfrac{1}{2}$. The vertical asymptote $x = -1$ of $f$ gives the open circle $(-1, 0)$: the graph comes as close as desired to this point, which does not belong to it.
Points. $f(x) = 1$ when $2x - 2 = x + 1$, that is $x = 3$; $f(x) = -1$ when $2x - 2 = -x - 1$, that is $x = \dfrac{1}{3}$: the invariant points are $(3, 1)$ and $\left(\dfrac{1}{3}, -1\right)$. The $y$-intercept is $\dfrac{1}{f(0)} = -\dfrac{1}{2}$.
Check. For $x \neq \pm 1$, $\dfrac{1}{f(x)} = \dfrac{x + 1}{2x - 2}$, a rational function whose asymptotes are $x = 1$ and $y = \dfrac{1}{2}$, as found; the graph of $y = \dfrac{1}{f(x)}$ is the graph of this rational function without its point $(-1, 0)$, where $f$ is not defined.

Example

Let $f(x) = x^3 - 4x + 1$. Use technology to find the vertical asymptotes and the turning points of the graph of $y = \dfrac{1}{f(x)}$, to $3$ significant figures, and sketch it.

Answer

Enter $Y_1 = x^3 - 4x + 1$ in the window $-3 \le x \le 3$, $-5 \le y \le 6$, and read its key features with the analysis tools (Method “Reading the Key Features of a Graph on the Screen”).

Vertical asymptotes. The tool zero displays $X = -2.1149075$, $X = 0.25410169$ and $X = 1.8608059$: the vertical asymptotes are $x \approx -2.11$, $x \approx 0.254$ and $x \approx 1.86$.
Turning points. The tool maximum displays $X = -1.1547005$, $Y = 4.0792014$, and the tool minimum $X = 1.1547005$, $Y = -2.0792014$. The local maximum of $f$ becomes a local minimum of $\dfrac{1}{f}$, with value $\dfrac{1}{4.0792014} \approx 0.245$, and the local minimum becomes a local maximum, with value $\dfrac{1}{-2.0792014} \approx -0.481$: the turning points are $(-1.15, 0.245)$ and $(1.15, -0.481)$.
Check on the screen. With $Y_2 = \dfrac{1}{Y_1}$, the tools minimum and maximum applied to $Y_2$ give the same two points. The calculator may join two branches across an asymptote with an almost vertical segment, which is not part of the graph.

Since $|f(x)|$ becomes as large as desired when $x$ becomes large, the $x$-axis is a horizontal asymptote; the four branches keep the sign of $f$.

Skills to practice

Recognising Features of the Graph of $y \equal \frac{1}{f(x)}$
Finding Asymptotes and Invariant Points
Studying $y \equal \frac{1}{f(x)}$ with Technology
Sketching the Graph of $y \equal \frac{1}{f(x)}$

IV Going further

Skills to practice

Reasoning about Transformations
Solving Problems with Transformations
Reasoning about Reciprocal Graphs
Solving Problems with Reciprocal Graphs
Quiz

\(x\)	\(0\)	\(1\)	\(3\)	\(4\)
\(f(x) = x^2 - 4\)	\(-4\)	\(-3\)	\(5\)	\(12\)
\(\dfrac{1}{f(x)}\)	\(-\dfrac{1}{4}\)	\(-\dfrac{1}{3}\)	\(\dfrac{1}{5}\)	\(\dfrac{1}{12}\)