Algebra-2

Let's analyze each function one by one based on the graph:

 

A.

1. Vertical asymptote:

To find the vertical asymptote, we set the denominator equal to zero. For , when , we get . So the vertical asymptote is at , which matches the given graph.

 

2. Horizontal asymptote:

Since the degree of the numerator and the denominator is the same (both are of degree ), we divide the leading coefficients. The leading coefficient of the numerator is and that of the denominator is also . So is the horizontal asymptote, which meets the graph.

 

3. Intercepts:

-intercept: Set , we have , which implies , so .

-intercept: Set , we get .

This function satisfies the graph.

 

B.

1. Vertical asymptotes:

First, factor the denominator . Set the denominator equal to zero, we get and as the vertical asymptotes. This does not match the requirement of having only one vertical asymptote at . Also, this function has a hole at because the factor can be canceled out (after considering the domain restrictions), which is against the given graph of having no hole. So this function is not correct.

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