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Reciprocal Functions

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Reciprocal Functions
 Let y=f(x)y=f(x)y=f(x) be a continuous polynomial function. The reciprocal function of y=f(x) y=f(x)~y=f(x)  is:
y=1f(x)\boxed{y=\frac{1}{f(x)}}y=f(x)1​​
with a vertical asymptote at f(x)=0f(x)=0f(x)=0 and a horizontal asymptote at y=0.y=0.y=0.

Vertical asymptotes are imaginary vertical lines that correspond to the zeros in the denominator of a reciprocal function. The function cannot touch or cross a vertical asymptote.

Horizontal asymptotes are imaginary horizontal lines that indicate the behavior of the function as x → ±∞x~\rightarrow~\pm\infinx → ±∞. The function can touch or cross a horizontal asymptote. 



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The Relationship Between y=f(x)  &  y=1f(x)\blue{y=f(x)~~\&~~y=\displaystyle\frac{1}{f(x)}}y=f(x)  &  y=f(x)1​:
Let's look at y=x  &  y=1xy=x~~\&~~y=\displaystyle\frac{1}{x}y=x  &  y=x1​ on the same grid.

         xy=xy=1x−3−3−1/3−2−2−1/2−1−1−100UND111221/2331/3\begin{array}{|c|c|c|}\hline
x&y=x&y=\frac{1}{x}\\\\\hline
-3&-3&-1/3\\\hline
-2&-2&-1/2\\\hline
-1&-1&-1\\\hline
0&0&\text{UND}\\\hline
1&1&1\\\hline
2&2&1/2\\\hline
3&3&1/3\\\hline
\end{array}x−3−2−10123​y=x−3−2−10123​y=x1​−1/3−1/2−1UND11/21/3​​  


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y=xy=1xX-Interceptsx=0NAEnd Behaviourx →∞,y →∞x →−∞,y →−∞x →∞,y →0x →−∞,y →−0Points of(1,1)(1,1)Intersection(−1,−1)(−1,−1)Increasing(−∞,∞)NADecreasingNA(−∞,0)∪(0,∞)Domain(−∞,∞)(−∞,0)∪(0,∞)Range(−∞,∞)(−∞,0)∪(0,∞)\begin{array}{|l|c|c|}\hline\\
&y=x&y=\frac{1}{x}\\ \\
\hline\\
\footnotesize\textbf{X-Intercepts}&x=0&\text{NA}\\\\\hline\\
\footnotesize\textbf{End Behaviour}&\begin{array}{c}x~\rightarrow\infin,y~\rightarrow\infin\\
x~\rightarrow-\infin,y~\rightarrow-\infin\end{array}&\begin{array}{c}x~\rightarrow\infin,y~\rightarrow0\\
x~\rightarrow-\infin,y~\rightarrow-0\end{array}\\\\\hline\\
\footnotesize\textbf{Points of}&(1,1)&(1,1)\\
\footnotesize\textbf{Intersection}&(-1,-1)&(-1,-1)\\\\\hline\\
\footnotesize\textbf{Increasing}&(-\infin,\infin)&\text{NA}\\\\\hline\\
\footnotesize\textbf{Decreasing}&\text{NA}&(-\infty,0)\cup(0,\infty)\\\\\hline\\
\footnotesize\textbf{Domain}&(-\infin,\infin)&(-\infty,0)\cup(0,\infty)\\\\\hline\\
\footnotesize\textbf{Range}&(-\infin,\infin)&(-\infty,0)\cup(0,\infty)\\\\\hline
\end{array}X-InterceptsEnd BehaviourPoints ofIntersectionIncreasingDecreasingDomainRange​y=xx=0x →∞,y →∞x →−∞,y →−∞​(1,1)(−1,−1)(−∞,∞)NA(−∞,∞)(−∞,∞)​y=x1​NAx →∞,y →0x →−∞,y →−0​(1,1)(−1,−1)NA(−∞,0)∪(0,∞)(−∞,0)∪(0,∞)(−∞,0)∪(0,∞)​​

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Graphing Reciprocal Functions

If given y=f(x)y=f(x)y=f(x), the following are steps are provided to illustrate how to graph y=1f(x)y=\displaystyle\frac{1}{f(x)}y=f(x)1​ :

Sketch y=f(x)y=f(x)y=f(x), stating the domain and range.
Identify the x-intercepts on y=f(x)y=f(x)y=f(x)
Identify the points of intersection between y=f(x)y=f(x)y=f(x) and y=±1y=\pm1y=±1 (critical points)
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes (dotted lines)
Draw the horizontal asymptote (dotted lines)
Start from the critical points and draw the graph towards the asymptotes.

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Example 1
Graph y=1x−2y=\displaystyle\frac{1}{x-2}y=x−21​

Step 1.
Sketch y=x−2,\green{y=x-2,}y=x−2,stating the domain and range

     xy−2−4−1−30−21−1203142\begin{array}{|c|c|}\hline
x&y\\\hline
-2&-4\\\hline
-1&-3\\\hline
0&-2\\\hline
1&-1\\\hline
2&0\\\hline
3&1\\\hline
4&2\\\hline
\end{array}x−2−101234​y−4−3−2−1012​​

Domain: (−∞,∞)(-\infin,\infin)(−∞,∞)

Range: (−∞,∞)(-\infin,\infin)(−∞,∞)

Step 2. 
Identify the x-intercepts on y=x−2\green{y=x-2}y=x−2

The x-intercept is at (2,0)(2,0)(2,0)

Step 3.
Identify the points of intersection between y=x−2\green{y=x-2}y=x−2 and y=±1\green{y=\pm1}y=±1 (critical points)

The points of intersection are (3,1)  &  (1,−1)(3, 1)~~\&~~(1, -1)(3,1)  &  (1,−1)

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Step 4.
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes

Step 5
Draw the horizontal asymptote

Step 6. 
Start from the critical points and draw the graph towards the asymptotes.

The end behaviour of y=x−2 y=x-2~y=x−2 :
x→∞,y→∞x→−∞,y−→∞x\rightarrow\infin,y\rightarrow\infin\newline{}
x\rightarrow-\infin,y-\rightarrow\infinx→∞,y→∞x→−∞,y−→∞ 


The end behaviour of y=1x−2y=\displaystyle\frac{1}{x-2}y=x−21​ is:
x→∞,y→−0x→−∞,y→0x\rightarrow\infin,y\rightarrow-0\newline{}
x\rightarrow-\infin,y\rightarrow0x→∞,y→−0x→−∞,y→0

  

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Example 2

Graph y=1x2y=\displaystyle\frac{1}{x^2}y=x21​

Step 1.
Sketch y=x2\green{y=x^2}y=x2, stating the domain and range. 

       xy−24−11001124\begin{array}{|c|c|}\hline
x&y\\\hline
-2&4\\\hline
-1&1\\\hline
0&0\\\hline
1&1\\\hline
2&4\\\hline
\end{array}x−2−1012​y41014​​

Domain: (−∞,∞)(-\infin, \infin)(−∞,∞)

Range: [0,∞)[0,\infin)[0,∞)

Step 2. 
Identify the x-intercepts on y=x2\green{y=x^2}y=x2

The x-intercept is at (0,0)(0,0)(0,0)

Step 3.
Identify the points of intersection between y=x2\green{y=x^2}y=x2 and y=±1\green{y=\pm1}y=±1 (critical points)

The points of intersection are (1,1)  &  (−1,1)(1, 1)~~\&~~(-1, 1)(1,1)  &  (−1,1)

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Step 4.
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes

Step 5
Draw the horizontal asymptote

Step 6. 
Start from the critical points and draw the graph towards the asymptotes.

The end behaviour of y=x2 y=x^2~y=x2 :
x→∞,y→∞x→−∞,y→∞x\rightarrow\infin,y\rightarrow\infin\newline{}
x\rightarrow-\infin,y\rightarrow\infinx→∞,y→∞x→−∞,y→∞ 

The end behaviour of y=1x2y=\dfrac{1}{x^2}y=x21​ is:
x→∞,y→0x→−∞,y→0x\rightarrow\infin,y\rightarrow0\newline{}
x\rightarrow-\infin,y\rightarrow0x→∞,y→0x→−∞,y→0

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Example: Graphing Reciprocal of a Linear Function 

a) Sketch y=2x+3y=2x+3y=2x+3, identifying the end behaviour & x-intercepts.

b) Use the information from part a) to sketch a graph of the reciprocal function y=12x+3y=\displaystyle\frac{1}{2x+3}y=2x+31​, identifying vertical asymptotes, horizontal asymptotes, & critical points 

Step 1.
Sketch y=2x+3\green{y=2x+3}y=2x+3
     xy−2−1−1103152739411\begin{array}{|c|c|}\hline
x&y\\\hline
-2&-1\\\hline
-1&1\\\hline
0&3\\\hline
1&5\\\hline
2&7\\\hline
3&9\\\hline
4&11\\\hline
\end{array}x−2−101234​y−11357911​​

Step 2. 
Identify the x-intercepts on y=2x+3\green{y=2x+3}y=2x+3

The x-intercept is at (−1.5,0)(-1.5,0)(−1.5,0)

Step 3.
Identify the points of intersection between y=2x+3\green{y=2x+3}y=2x+3 and y=±1\green{y=\pm1}y=±1 (critical points)

The points of intersection are (−1,1)  &  (−2,−1)(-1, 1)~~\&~~(-2, -1)(−1,1)  &  (−2,−1)

Step 4.
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes

Vertical Asymptote: x=−32x=-\displaystyle\frac{3}{2}x=−23​

Step 5
Draw the horizontal asymptote y = 0

Step 6. 
Start from the critical points and draw the graph towards the asymptotes.

The end behaviour of y=2x+3 y=2x+3~y=2x+3 :
x→∞,y→∞x→−∞,y−→∞x\rightarrow\infin,y\rightarrow\infin\newline{}
x\rightarrow-\infin,y-\rightarrow\infinx→∞,y→∞x→−∞,y−→∞ 

The end behaviour of y=12x+3y=\displaystyle\frac{1}{2x+3}y=2x+31​ is:
x→∞,y→0x→−∞,y→0x\rightarrow\infin,y\rightarrow0\newline{}
x\rightarrow-\infin,y\rightarrow0x→∞,y→0x→−∞,y→0

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Example: Graphing Reciprocal of a Quadratic Function
a) Sketch y=x2−9y=x^2-9y=x2−9, identifying the end behaviour, x-intercepts, y-intercepts, & any minimums and maximums.

b) Use the information from part a) to sketch a graph of the reciprocal function y=1x2−9y=\displaystyle\frac{1}{x^2-9}y=x2−91​, identifying vertical asymptotes, horizontal asymptotes, minimums/maximums, & critical points 

Step 1.
Sketch y=x2−9\green{y=x^2-9}y=x2−9

     xy−39−24−1100112439\begin{array}{|c|c|}\hline
x&y\\\hline
-3&9\\\hline
-2&4\\\hline
-1&1\\\hline
0&0\\\hline
1&1\\\hline
2&4\\\hline
3&9\\\hline
\end{array}x−3−2−10123​y9410149​​

Step 2. 
Identify the x-intercepts, y-intercepts, & min/max on y=x2−9\green{y=x^2-9}y=x2−9

The x-intercept is at (−3,0)  &  (3,0)(-3,0)~~\&~~(3, 0)(−3,0)  &  (3,0)

The y-intercept is at (0,−9)(0,-9)(0,−9)

The minimum is at (0,−9)(0,-9)(0,−9)

Step 3.
Identify the points of intersection between y=x2−9\green{y=x^2-9}y=x2−9 and y=±1\green{y=\pm1}y=±1 (critical points)

The points of intersection are (±10,1)  &  (±8,−1)(\pm\sqrt{10},1)~~\&~~(\pm\sqrt{8},-1)(±10​,1)  &  (±8​,−1)

Step 4.
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes

Vertical Asymptotes: x=±3x=\pm3x=±3

Step 5
Draw the horizontal asymptote y = 0

Step 6. 
Start from the critical points and draw the graph towards the asymptotes.

The minimum (0,−9)(0,-9)(0,−9) turns into a maximum (0,−19)(0,-\frac{1}{9})(0,−91​)

The end behaviour of y=x2−9 y=x^2-9~y=x2−9 :
x→∞,y→∞x→−∞,y→∞x\rightarrow\infin,y\rightarrow\infin\newline{}
x\rightarrow-\infin,y\rightarrow\infinx→∞,y→∞x→−∞,y→∞ 

The end behaviour of y=1x2−9y=\displaystyle\frac{1}{x^2-9}y=x2−91​ is:
x→∞,y→0x→−∞,y→0x\rightarrow\infin,y\rightarrow0\newline{}
x\rightarrow-\infin,y\rightarrow0x→∞,y→0x→−∞,y→0

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Example: Graphing Reciprocal of a Quadratic Function
a) Sketch y=x2+4y=x^2+4y=x2+4, identifying the end behaviour & x-intercepts.

b) Use the information from part a) to sketch a graph of the reciprocal function y=1x2+4y=\displaystyle\frac{1}{x^2+4}y=x2+41​, identifying vertical asymptotes, horizontal asymptotes, & critical points 

Step 1.
Sketch y=x2+4\green{y=x^2+4}y=x2+4

     xy−28−15041528\begin{array}{|c|c|}\hline
x&y\\\hline
-2&8\\\hline
-1&5\\\hline
0&4\\\hline
1&5\\\hline
2&8\\\hline
\end{array}x−2−1012​y85458​​

Step 2. 
Identify the x-intercepts on y=x2+4\green{y=x^2+4}y=x2+4

There are no x-intercepts.

Step 3.
Identify the points of intersection between y=x2+4\green{y=x^2+4}y=x2+4 and y=±1\green{y=\pm1}y=±1 (critical points)

There are no points of intersection.

Step 4.
Turn the x-intercepts into vertical asymptotes & draw the vertical asymptotes

There are no vertical asymptotes.

Step 5
Draw the horizontal asymptote y = 0

Step 6. 
Start from the critical points and draw the graph towards the asymptotes.

The minimum (0,4)(0,4)(0,4) turns into a maximum (0,14)(0,\frac{1}{4})(0,41​)

The end behaviour of y=x2+4 y=x^2+4~y=x2+4 :
x→∞,y→∞x→−∞,y→∞x\rightarrow\infin,y\rightarrow\infin\newline{}
x\rightarrow-\infin,y\rightarrow\infinx→∞,y→∞x→−∞,y→∞ 

The end behaviour of y=1x2+4y=\displaystyle\frac{1}{x^2+4}y=x2+41​ is:
x→∞,y→0x→−∞,y→0x\rightarrow\infin,y\rightarrow0\newline{}
x\rightarrow-\infin,y\rightarrow0x→∞,y→0x→−∞,y→0

Practice: Graphing Reciprocal Functions
Sketch a graph of y=(x−3)2−4y=(x-3)^2-4y=(x−3)2−4 and it's reciprocal on the same grid, labeling all asymptotes & minimum/maximums points. 

Practice: Graphing Reciprocal Functions
Determine the equation for the function graphed below:

f(x)=

I don't know

Extra Practice

Rational Functions

Identify the information for the following reciprocal function:
y=1x−2y=\dfrac{1}{x-2}y=x−21​

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Determine the equation for the function graphed below:

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Determine the equation for the function graphed below:

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Sketch a graph of y=−2x2+4y=-2x^2+4y=−2x2+4 and it's reciprocal on the same grid, labeling vertical asymptotes.

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Sketch a graph of the reciprocal ofy=(x+2)2−2y=(x+2)^2-2y=(x+2)2−2 .

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Sketch a graph of the reciprocal of the function below:

Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions
Sketch a graph of the reciprocal of the function below:

Wize High School Grade 11 Math Textbook > Reciprocal Functions

Reciprocal Functions

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Reciprocal Functions

The Relationship Between y=f(x) & y=1f(x)\blue{y=f(x)~~\&~~y=\displaystyle\frac{1}{f(x)}}y=f(x) & y=f(x)1​:

Graphing Reciprocal Functions

Example: Graphing Reciprocal of a Linear Function

Example: Graphing Reciprocal of a Quadratic Function

Example: Graphing Reciprocal of a Quadratic Function

Practice: Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions

Practice: Graphing Reciprocal Functions

Extra Practice

Rational Functions

Graphing Reciprocal Functions

Graphing Reciprocal Functions

Graphing Reciprocal Functions

Graphing Reciprocal Functions

Graphing Reciprocal Functions

Graphing Reciprocal Functions

The Relationship Between $\blue{y=f(x)\&y=\displaystyle\frac{1}{f(x)}}$ :