Question

find an equation of a rational function f that satisfies the given conditions. vertical asymptotes: x = -1, x = 0 horizontal asymptote: y = 0 x - intercept: 3; f(4) = 1 f(x) =

Explanation:

Step1: Set up the general form of the rational function

A rational function with vertical asymptotes at $x = - 1$ and $x=0$ has factors of $(x + 1)$ and $x$ in the denominator. So, $f(x)=\frac{a(x - b)}{x(x + 1)}$. Since the $x$-intercept is 3, then $b = 3$, and the function becomes $f(x)=\frac{a(x - 3)}{x(x + 1)}$.

Step2: Use the given point to find the value of $a$

We know that $f(4)=1$. Substitute $x = 4$ into $f(x)=\frac{a(x - 3)}{x(x + 1)}$:
\[1=\frac{a(4 - 3)}{4(4 + 1)}\]
\[1=\frac{a\times1}{4\times5}\]
\[1=\frac{a}{20}\]
\[a = 20\]

Answer:

$f(x)=\frac{20(x - 3)}{x(x + 1)}$