Question 1104537
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a. A rational function that has a vertical asymptote at -5 and a hole at 7.<br>
The graph of a rational function has a hole at x=a if there are linear factors of (x-a) in both numerator and denominator.<br>
The graph of a rational function has a vertical asymptote at x=a if there is a linear factor of x-a in the denominator but not in the numerator.<br>
So to get a vertical asymptote at x=-5 there must be a factor of (x+5) in the denominator but not in the numerator; to get a hole at x=7 there must be a factor of (x-7) in both numerator and denominator .<br>
The simplest rational function with those factors is
{{{(x-7)/((x+5)(x-7))}}}<br>
b. A rational function that has no vertical asymptote and does have a slant asymptote<br>
The graph of a rational function has a slant asymptote if the degree of the numerator is 1 greater than the degree of the denominator.<br>
Assuming for simplicity that there are no holes in the graph, caused by identical linear factors in both numerator and denominator, then if the graph of a rational function has NO vertical asymptotes, then there are no linear factors in the denominator.  That means the denominator has no real zeros.<br>
So a rational function that has a slant asymptote but no vertical asymptote must have a denominator with no real zeros, and a numerator with degree 1 greater than the degree of the denominator.<br>
A simple example of such a function is
{{{x^3/(x^2+1)}}}