select the correct answer. rational function h is continuous, with a horizontal asymptote at y = 1. which…

Explanation:

Step1: Recall horizontal - asymptote rules for rational functions

For a rational function $h(x)=\frac{f(x)}{g(x)}=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, if $n = m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n\lt m$, the horizontal asymptote is $y = 0$; if $n>m$, there is no horizontal asymptote.

Step2: Analyze option A

For $h(x)=\frac{x^2 - 16}{x^2+16}$, here $n = m=2$, $a_n = 1$, $b_m = 1$. So the horizontal asymptote is $y=\frac{1}{1}=1$. Also, the denominator $x^2 + 16$ is never zero since $x^2\geq0$ for all real $x$, so $x^2+16\geq16>0$ and the function is continuous.

Step3: Analyze option B

For $h(x)=\frac{x^2 + 16}{x^2-16}$, the denominator $x^2 - 16=(x + 4)(x - 4)=0$ when $x=\pm4$, so the function is not continuous.

Step4: Analyze option C

For $h(x)=\frac{x^2 - 16}{x - 4}=\frac{(x + 4)(x - 4)}{x - 4}=x + 4,x\neq4$. It has a removable discontinuity at $x = 4$ and no horizontal asymptote.

Step5: Analyze option D

For $h(x)=\frac{x + 4}{x^2+16}$, since $n = 1$ and $m=2$ ($n\lt m$), the horizontal asymptote is $y = 0$.

Answer:

A. $h(x)=\frac{x^2 - 16}{x^2+16}$