Question

1. write a function that has zeros at x = -3 and x = 2, and vertical asymptotes at x = -5 and x = 7. it has a horizontal asymptote of y = 1.

Explanation:

Step1: Form the numerator for zeros

A rational - function has zeros at the values of \(x\) that make the numerator equal to zero. If the zeros are at \(x=-3\) and \(x = 2\), then the numerator of the rational function is \(f(x)=(x + 3)(x - 2)=x^{2}+x - 6\).

Step2: Form the denominator for vertical asymptotes

A rational function has vertical asymptotes at the values of \(x\) that make the denominator equal to zero. If the vertical asymptotes are at \(x=-5\) and \(x = 7\), then the denominator of the rational function is \(g(x)=(x + 5)(x - 7)=x^{2}-2x-35\).

Step3: Adjust for horizontal asymptote

For a rational function \(\frac{f(x)}{g(x)}\) where \(f(x)=a_nx^n+\cdots\) and \(g(x)=b_mx^m+\cdots\), if \(n = m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\). Since the horizontal asymptote is \(y = 1\) and the degrees of the numerator and denominator are the same (both are 2), we can use the function \(y=\frac{(x + 3)(x - 2)}{(x + 5)(x - 7)}=\frac{x^{2}+x - 6}{x^{2}-2x-35}\).

Answer:

\(y=\frac{(x + 3)(x - 2)}{(x + 5)(x - 7)}\)