Question

how would i answer and what would be the answer?

Answer 1

Given:

$f(x)=-\frac{1}{x-2}+3$

We need to find asymptotes of the given function.

Recall that a horizontal asymptote is a horizontal line, y=a, that has the property that either:

$\lim _{x\to\infty}f(x)=a$

or

$\lim _{x\to-\infty}f(x)=a$

Taking the limit of the given function, we get

$\lim _{x\to\infty}f(x)=\lim _{x\to\infty}(-\frac{1}{x-2}+3)$

$\lim _{x\to\infty}f(x)=0+3$

$\lim _{x\to\infty}f(x)=3$

The horizontal asymptote is y=3.

Recall that a vertical asymptote is a vertical line, x=a, that has the property that either:

$\lim _{x\to a^-}f(x)=\pm\infty$

or

$\lim _{x\to a^+}f(x)=\pm\infty$

Taking limit to the given function, we get

$\lim _{x\to a^+}f(x)=\lim _{x\to a^+}(-\frac{1}{x-2}+3)$

If we take a=2, the limit will be infinity.

$\lim _{x\to2^+}f(x)=\lim _{x\to2^+}(-\frac{1}{x-2}+3)=\infty$

Hence the vertical asymptotes x=2.

From these two asymptotes, we can say that the first option or third option would be the graph.

Consider the point (3,2) from the first option graph.

Substitute x=3 and f(3)=2 in the given function, we get

$2=-\frac{1}{3-2}+3$

$2=-1+3$$2=2$

The point (3,2) satisfies the given function.

Hence the graph of the function is