<< Chapter < Page | Chapter >> Page > |
Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x -intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x -intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.
If a rational function has x -intercepts at $\text{\hspace{0.17em}}x={x}_{1},{x}_{2},\mathrm{...},{x}_{n},\text{\hspace{0.17em}}$ vertical asymptotes at $\text{\hspace{0.17em}}x={v}_{1},{v}_{2},\dots ,{v}_{m},\text{\hspace{0.17em}}$ and no $\text{\hspace{0.17em}}{x}_{i}=\text{any}{v}_{j},\text{\hspace{0.17em}}$ then the function can be written in the form:
where the powers $\text{\hspace{0.17em}}{p}_{i}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}{q}_{i}\text{\hspace{0.17em}}$ on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ can be determined given a value of the function other than the x -intercept or by the horizontal asymptote if it is nonzero.
Given a graph of a rational function, write the function.
Write an equation for the rational function shown in [link] .
The graph appears to have x -intercepts at $\text{\hspace{0.17em}}x=\u20132\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=3.\text{\hspace{0.17em}}$ At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at $\text{\hspace{0.17em}}x=\u20131\text{\hspace{0.17em}}$ seems to exhibit the basic behavior similar to $\text{\hspace{0.17em}}\frac{1}{x},\text{\hspace{0.17em}}$ with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ is exhibiting a behavior similar to $\text{\hspace{0.17em}}\frac{1}{{x}^{2}},\text{\hspace{0.17em}}$ with the graph heading toward negative infinity on both sides of the asymptote. See [link] .
We can use this information to write a function of the form
To find the stretch factor, we can use another clear point on the graph, such as the y -intercept $\text{\hspace{0.17em}}(0,\mathrm{\u20132}).$
This gives us a final function of $\text{\hspace{0.17em}}f(x)=\frac{4(x+2)(x-3)}{3(x+1){(x-2)}^{2}}.$
Access these online resources for additional instruction and practice with rational functions.
Notification Switch
Would you like to follow the 'Precalculus' conversation and receive update notifications?