# 5.6 Rational functions  (Page 7/16)

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Find the vertical and horizontal asymptotes of the function:

$f\left(x\right)=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-2\right)\left(x+3\right)}$

Vertical asymptotes at $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=–3;\text{\hspace{0.17em}}$ horizontal asymptote at $\text{\hspace{0.17em}}y=4.$

## Intercepts of rational functions

A rational function    will have a y -intercept at $\text{\hspace{0.17em}}f\left(0\right)$ , if the function is defined at zero. A rational function will not have a y -intercept if the function is not defined at zero.

Likewise, a rational function will have x -intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x -intercepts can only occur when the numerator of the rational function is equal to zero.

## Finding the intercepts of a rational function

Find the intercepts of $\text{\hspace{0.17em}}f\left(x\right)=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-1\right)\left(x+2\right)\left(x-5\right)}.$

We can find the y -intercept by evaluating the function at zero

$\begin{array}{ccc}\hfill f\left(0\right)& =& \frac{\left(0-2\right)\left(0+3\right)}{\left(0-1\right)\left(0+2\right)\left(0-5\right)}\hfill \\ & =& \frac{-6}{10}\hfill \\ & =& -\frac{3}{5}\hfill \\ & =& -0.6\hfill \end{array}$

The x -intercepts will occur when the function is equal to zero:

The y -intercept is $\text{\hspace{0.17em}}\left(0,–0.6\right),\text{\hspace{0.17em}}$ the x -intercepts are $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(–3,0\right).\text{\hspace{0.17em}}$ See [link] .

Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x - and y -intercepts and the horizontal and vertical asymptotes.

For the transformed reciprocal squared function, we find the rational form. $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{{\left(x-3\right)}^{2}}-4=\frac{1-4{\left(x-3\right)}^{2}}{{\left(x-3\right)}^{2}}=\frac{1-4\left({x}^{2}-6x+9\right)}{\left(x-3\right)\left(x-3\right)}=\frac{-4{x}^{2}+24x-35}{{x}^{2}-6x+9}$

Because the numerator is the same degree as the denominator we know that as is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ because as $\text{\hspace{0.17em}}x\to 3,f\left(x\right)\to \infty .\text{\hspace{0.17em}}$ We then set the numerator equal to 0 and find the x -intercepts are at $\text{\hspace{0.17em}}\left(2.5,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(3.5,0\right).\text{\hspace{0.17em}}$ Finally, we evaluate the function at 0 and find the y -intercept to be at $\text{\hspace{0.17em}}\left(0,\frac{-35}{9}\right).$

## Graphing rational functions

In [link] , we see that the numerator of a rational function reveals the x -intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See [link] .

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See [link] .

For example, the graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{{\left(x+1\right)}^{2}\left(x-3\right)}{{\left(x+3\right)}^{2}\left(x-2\right)}\text{\hspace{0.17em}}$ is shown in [link] .

• At the x -intercept $\text{\hspace{0.17em}}x=-1\text{\hspace{0.17em}}$ corresponding to the $\text{\hspace{0.17em}}{\left(x+1\right)}^{2}\text{\hspace{0.17em}}$ factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.
• At the x -intercept $\text{\hspace{0.17em}}x=3\text{\hspace{0.17em}}$ corresponding to the $\text{\hspace{0.17em}}\left(x-3\right)\text{\hspace{0.17em}}$ factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
• At the vertical asymptote $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ corresponding to the $\text{\hspace{0.17em}}{\left(x+3\right)}^{2}\text{\hspace{0.17em}}$ factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{{x}^{2}}.$
• At the vertical asymptote $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ corresponding to the $\text{\hspace{0.17em}}\left(x-2\right)\text{\hspace{0.17em}}$ factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}.$

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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Abhi
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salma
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salma
Commplementary angles
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Sherica
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Sherica
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Tamia
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Uday
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salma
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a perfect square v²+2v+_
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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