Algebra - Rational Functions

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Section 4.8 : Rational Functions

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1. Sketch the graph of the following function. Clearly identify all intercepts and asymptotes.

$f\left( x \right) = \frac{{ - 4}}{{x - 2}}$

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Let’s first find the intercepts for this function.

The $y$ -intercept is the point $\left( {0,f\left( 0 \right)} \right) = \left( {0,2} \right)$ .

For the $x$ -intercepts we set the numerator equal to zero and solve. However, in this case the numerator is a constant (-4 specifically) and so can’t ever be zero. Therefore, this function will have no $x$ -intercepts.

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We can find any vertical asymptotes be setting the denominator equal to zero and solving. Doing that for this function gives,

$x - 2 = 0\hspace{0.25in} \to \hspace{0.25in}x = 2$

So, we’ll have a vertical asymptote at $x = 2$ .

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For this equation the largest exponent of $x$ in the numerator is zero since the numerator is a constant. The largest exponent of $x$ in the denominator is 1, which is larger than the largest exponent in the numerator, and so the $x$ -axis will be the horizontal asymptote.

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From Step 2 we saw we only have one vertical asymptote and so we only have two regions to our graph : $x < 2$ and $x > 2$ .

We’ll need a point in each region to determine if it will be above or below the horizontal asymptote. Here are a couple of function evaluations for the points.

$\begin{align*}f\left( 0 \right) & = 2 & \to \hspace{0.25in} & \left( {0,2} \right)\\ f\left( 3 \right) & = - 4 & \to \hspace{0.25in} & \left( {3, - 4} \right)\end{align*}$

Note that the first evaluation didn’t really need to be done since it was just the $y$ -intercept which we had already found in the first step. It was included here mostly for the sake of completeness.

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Here is a sketch of the function with the points found above. The vertical asymptote is indicated with a blue dashed line and recall that the horizontal asymptote is just the $x$ -axis.