We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs and notice some of their features.

Several things are apparent if we examine the graph of [latex]f\left(x\right)=\dfrac{1}{x}[/latex].

1. On the left branch of the graph, the curve approaches the [latex]x[/latex]-axis [latex]\left(y=0\right) \text{ as } x\to -\infty [/latex].
2. As the graph approaches [latex]x=0[/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.
3. Finally, on the right branch of the graph, the curves approaches the [latex]x[/latex]–axis [latex]\left(y=0\right) \text{ as } x\to \infty [/latex].

To summarize, we use arrow notation to show that [latex]x[/latex] or [latex]f\left(x\right)[/latex] is approaching a particular value.

Local Behavior of [latex]f\left(x\right)=\frac{1}{x}[/latex]

Let’s begin by looking at the reciprocal function, [latex]f\left(x\right)=\frac{1}{x}[/latex]. We cannot divide by zero, which means the function is undefined at [latex]x=0[/latex]; so zero is not in the domain.

As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.

We write in arrow notation:

[latex]\text{as }x\to {0}^{-},f\left(x\right)\to -\infty [/latex]

As [latex]x[/latex] approaches [latex]0[/latex] from the left (negative) side, [latex]f(x)[/latex] will approach negative infinity.

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.

We write in arrow notation:

[latex]\text{As }x\to {0}^{+}, f\left(x\right)\to \infty [/latex].

As [latex]x[/latex] approaches [latex]0[/latex] from the right (positive) side, [latex]f(x)[/latex] will approach infinity.

This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line [latex]x=0[/latex] as the input becomes close to zero.

End Behavior of [latex]f\left(x\right)=\frac{1}{x}[/latex]

As the values of [latex]x[/latex] approach infinity, the function values approach 0. As the values of [latex]x[/latex] approach negative infinity, the function values approach 0. Symbolically, using arrow notation

[latex]\text{As }x\to \infty ,f\left(x\right)\to 0,\text{and as }x\to -\infty ,f\left(x\right)\to 0[/latex].

Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line [latex]y=0[/latex].

Example: Using Arrow Notation

Use arrow notation to describe the end behavior and local behavior of the function below.

Show Solution

Notice that the graph is showing a vertical asymptote at [latex]x=2[/latex], which tells us that the function is undefined at [latex]x=2[/latex].

As [latex]x\to {2}^{-},\hspace{2mm}f\left(x\right)\to -\infty[/latex], and as [latex]x\to {2}^{+},\text{ }f\left(x\right)\to \infty [/latex]

And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[/latex]. As the inputs increase without bound, the graph levels off at 4.

As [latex]x\to \infty ,\text{ }f\left(x\right)\to 4[/latex], and as [latex]x\to -\infty ,\text{ }f\left(x\right)\to 4[/latex]

Example: Using Transformations to Graph a Rational Function

Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

Show Solution

Shifting the graph left 2 and up 3 would result in the function

[latex]f\left(x\right)=\dfrac{1}{x+2}+3[/latex]

or equivalently, by giving the terms a common denominator,

[latex]f\left(x\right)=\dfrac{3x+7}{x+2}[/latex]

The graph of the shifted function is displayed below.

Notice that this function is undefined at [latex]x=-2[/latex], and the graph also is showing a vertical asymptote at [latex]x=-2[/latex].

As [latex]x\to -{2}^{-}, f\left(x\right)\to -\infty[/latex] , and as  [latex]x\to -{2}^{+}, f\left(x\right)\to \infty [/latex]

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[/latex].

[latex]\text{As }x\to \pm \infty , f\left(x\right)\to 3[/latex]

Analysis of the Solution

Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.

Try It

Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

Show Solution

The function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\to 3,f\left(x\right)\to \infty[/latex], and as [latex]x\to \pm \infty ,f\left(x\right)\to -4[/latex].

The function is [latex]f\left(x\right)=\dfrac{1}{{\left(x - 3\right)}^{2}}-4[/latex].

A Mixing Problem

In the previous example, we shifted a toolkit function in a way that resulted in the function [latex]f\left(x\right)=\dfrac{3x+7}{x+2}[/latex]. This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

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