the graph of which of the following functions in the xy - plane has at least one x - intercept, at least one…

Answer:

A. $f(x)=\frac{x^{2}-16}{x^{2}-x - 6}$

Explanation:

Step1: Factor the numerator and denominator

$x^{2}-16=(x + 4)(x - 4)$; $x^{2}-x - 6=(x - 3)(x+2)$

Step2: Find x - intercepts

Set $y = 0$, then $(x + 4)(x - 4)=0$, so $x=-4$ or $x = 4$. There are x - intercepts.

Step3: Find holes

There are no common factors in the numerator and denominator to create a hole.

Step4: Find vertical asymptotes

Set the denominator $(x - 3)(x + 2)=0$, we get $x=3$ and $x=-2$, so there are vertical asymptotes.

Step5: Find horizontal asymptote

Since the degree of the numerator and denominator are the same (both 2), the horizontal asymptote is $y=\frac{1}{1}=1$ (the ratio of the leading - coefficients).

For option B:

Step1: Factor the numerator and denominator

$x^{2}-16=(x + 4)(x - 4)$; $x^{2}-x - 30=(x - 6)(x+5)$ There are x - intercepts at $x = 4$ and $x=-4$, vertical asymptotes at $x = 6$ and $x=-5$, and horizontal asymptote $y = 1$, but no holes.

For option C:

Step1: Factor the numerator and denominator

$x^{2}-4=(x + 2)(x - 2)$; $x^{2}-x - 30=(x - 6)(x+5)$ There are x - intercepts at $x = 2$ and $x=-2$, vertical asymptotes at $x = 6$ and $x=-5$, and horizontal asymptote $y = 1$, but no holes.

For option D:

Step1: Factor the numerator and denominator

$x^{2}-4=(x + 2)(x - 2)$; $x^{2}-x - 6=(x - 3)(x+2)$ There is a hole at $x=-2$ (common factor), x - intercepts at $x = 2$, vertical asymptote at $x = 3$, and horizontal asymptote $y = 1$. But we want a function with at least one of each feature, and this one has only one x - intercept.

So the function $f(x)=\frac{x^{2}-16}{x^{2}-x - 6}$ satisfies the conditions of having at least one x - intercept, at least one hole, at least one vertical asymptote, and a horizontal asymptote.