Question

sketch the graph of the following function. indicate where the function is increasing or decreasing, where any relative extrema occur, where asymptotes occur, where the graph is concave up or concave down, where any points of inflection occur, and where any intercepts occur.
f(x) = \frac{x + 1}{x^{2}-8x - 9}

the coordinates of the relative extrema are
(type an ordered - pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

there are no relative extrema.

determine the vertical asymptote(s). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

the function has two vertical asymptotes. the leftmost asymptote is and the rightmost asymptote is
(type equations.)

the function has one vertical asymptote, x = 9. (type an equation.)

the function has no vertical asymptotes.

determine the horizontal asymptote(s). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

the function has one horizontal asymptote, (type an equation.)

the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is
(type equations.)

the function has no horizontal asymptotes.

Explanation:

Step1: Factor the denominator

First, factor \(x^{2}-8x - 9=(x - 9)(x+1)\). So the function is \(f(x)=\frac{x + 1}{(x - 9)(x + 1)}\), but we need to consider the domain restrictions. The domain is all real - numbers except \(x=9\) and \(x=-1\). After canceling out the common factor \((x + 1)\) (for \(x
eq - 1\)), \(f(x)=\frac{1}{x - 9},x
eq - 1\).

Step2: Find vertical asymptotes

Vertical asymptotes occur where the denominator of the simplified function is zero. Set \(x-9 = 0\), we get \(x = 9\).

Step3: Find horizontal asymptotes

For a rational function of the form \(y=\frac{a}{x - h}+k\) (in our case \(a = 1\), \(h = 9\), \(k = 0\)), as \(x\to\pm\infty\), \(y\to0\). So the horizontal asymptote is \(y = 0\).

Step4: Check for relative extrema

Take the derivative of \(y=\frac{1}{x - 9}=(x - 9)^{-1}\). Using the power - rule \((u^{n})^\prime=nu^{n - 1}u^\prime\), where \(u=x - 9\) and \(n=-1\), we have \(y^\prime=-(x - 9)^{-2}=-\frac{1}{(x - 9)^{2}}\). Since \(y^\prime\) is never zero (the numerator of \(y^\prime\) is non - zero for all real \(x
eq9\)), there are no relative extrema.

Answer:

1. B. There are no relative extrema
2. B. The function has one vertical asymptote, \(x = 9\)
3. A. The function has one horizontal asymptote, \(y = 0\)