Question

find an equation of a rational function f that satisfies the given conditions. vertical asymptotes: x = -3, x = 4 horizontal asymptote: y = 2 x - intercept: -5, 2; hole at x = 0 f(x) =

Explanation:

Step1: Determine the denominator for vertical asymptotes

For vertical asymptotes at $x = - 3$ and $x = 4$, the denominator has factors $(x + 3)$ and $(x - 4)$. So far, the denominator $q(x)=(x + 3)(x - 4)=x^{2}-x - 12$.

Step2: Determine the numerator for x - intercepts

For $x$-intercepts at $x=-5$ and $x = 2$, the numerator has factors $(x + 5)$ and $(x - 2)$. Also, since there is a hole at $x = 0$, both the numerator and denominator have a factor of $x$. So the numerator $p(x)=ax(x + 5)(x - 2)=ax(x^{2}+3x - 10)=ax^{3}+3ax^{2}-10ax$.

Step3: Determine the value of a for horizontal asymptote

The horizontal asymptote is $y = 2$. Since the degrees of the numerator and denominator are the same (both degree 3), the ratio of the leading - coefficients gives the horizontal asymptote. The leading - coefficient of the denominator is 1 and of the numerator is $a$. So $a = 2$.

Answer:

$f(x)=\frac{2x(x + 5)(x - 2)}{x(x + 3)(x - 4)}=\frac{2x^{3}+6x^{2}-20x}{x^{3}-x^{2}-12x}$