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}-6x - 9}
determine the horizontal asymptote(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the horizontal asymptote(s) is(are)
(type an equation. use a comma to separate answers as needed.)
b. the function has no horizontal asymptotes.
determine the slant asymptote(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the slant asymptote(s) is(are)
(type an equation. use a comma to separate answers as needed.)
b. the function has no slant asymptotes.
on what interval(s) is ( f ) concave up and on what interval(s) is ( f ) concave down? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. the function is concave down on and is never concave up.
(simplify your answer. type your answer in interval - notation. use a comma to separate answers as needed.)
b. the function is concave up on and concave down on
(simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.)
c. the function is concave up on and is never concave down.
(simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.)

Explanation:

Step1: Simplify the function

First, factor the denominator of \(f(x)=\frac{x + 1}{x^{2}-6x - 9}\). The denominator \(x^{2}-6x - 9\) cannot be factored easily using simple integer - factoring methods.

Step2: Find vertical asymptotes

Set the denominator equal to zero: \(x^{2}-6x - 9=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-6\), \(c=-9\)), we have \(x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-9)}}{2\times1}=\frac{6\pm\sqrt{36 + 36}}{2}=\frac{6\pm\sqrt{72}}{2}=\frac{6\pm6\sqrt{2}}{2}=3\pm3\sqrt{2}\). So the vertical asymptotes are \(x = 3 + 3\sqrt{2}\) and \(x=3 - 3\sqrt{2}\).

Step3: Find horizontal asymptotes

Since the degree of the numerator (\(n = 1\)) is less than the degree of the denominator (\(m = 2\)), the horizontal asymptote is \(y = 0\).

Step4: Find the first - derivative

Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u=x + 1\), \(u^\prime=1\), \(v=x^{2}-6x - 9\), \(v^\prime=2x-6\). Then \(f^\prime(x)=\frac{1\times(x^{2}-6x - 9)-(x + 1)\times(2x - 6)}{(x^{2}-6x - 9)^{2}}=\frac{x^{2}-6x - 9-(2x^{2}-6x+2x - 6)}{(x^{2}-6x - 9)^{2}}=\frac{x^{2}-6x - 9 - 2x^{2}+4x + 6}{(x^{2}-6x - 9)^{2}}=\frac{-x^{2}-2x - 3}{(x^{2}-6x - 9)^{2}}\).

Step5: Find the second - derivative

Using the quotient rule again on \(y = f^\prime(x)=\frac{-x^{2}-2x - 3}{(x^{2}-6x - 9)^{2}}\). Let \(u=-x^{2}-2x - 3\), \(u^\prime=-2x - 2\), \(v=(x^{2}-6x - 9)^{2}\), \(v^\prime = 2(x^{2}-6x - 9)(2x - 6)\). Then \(f^{\prime\prime}(x)=\frac{(-2x - 2)(x^{2}-6x - 9)^{2}-(-x^{2}-2x - 3)\times2(x^{2}-6x - 9)(2x - 6)}{(x^{2}-6x - 9)^{4}}=\frac{(-2x - 2)(x^{2}-6x - 9)-2(-x^{2}-2x - 3)(2x - 6)}{(x^{2}-6x - 9)^{3}}\). Expand and simplify:

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Set \(f^{\prime\prime}(x) = 0\) to find inflection points. \(2x^{3}+6x^{2}+18x - 18 = 0\), or \(x^{3}+3x^{2}+9x - 9=0\). This cubic equation is difficult to solve exactly. We can use a graphing utility or test intervals.
We can also analyze the sign of \(f^{\prime\prime}(x)\) by choosing test points in the intervals separated by the vertical asymptotes \(x=3 - 3\sqrt{2}\approx - 1.24\) and \(x=3 + 3\sqrt{2}\approx7.24\).
Let's choose \(x=-2\), \(x = 0\), and \(x = 8\) as test - points.
For \(x=-2\): \(f^{\prime\prime}(-2)=\frac{2\times(-2)^{3}+6\times(-2)^{2}+18\times(-2)-18}{((-2)^{2}-6\times(-2)-9)^{3}}=\frac{-16 + 24-36 - 18}{(4 + 12-9)^{3}}=\frac{-46}{343}<0\).
For \(x = 0\): \(f^{\prime\prime}(0)=\frac{2\times0^{3}+6\times0^{2}+18\times0-18}{(0^{2}-6\times0-9)^{3}}=\frac{-18}{-729}=\frac{2}{81}>0\).
For \(x = 8\): \(f^{\prime\prime}(8)=\frac{2\times8^{3}+6\times8^{2}+18\times8-18}{(8^{2}-6\times8-9)^{3}}=\frac{1024+384 + 144-18}{(64-48 - 9)^{3}}=\frac{1534}{512}>0\).
The function is concave down on \((-\infty,3 - 3\sqrt{2})\) and concave up on \((3 - 3\sqrt{2},3 + 3\sqrt{2})\cup(3 + 3\sqrt{2},\infty)\).

Answer:

Vertical asymptotes: \(x = 3 + 3\sqrt{2}\), \(x=3 - 3\sqrt{2}\); Horizontal asymptote: \(y = 0\); Concave down on \((-\infty,3 - 3\sqrt{2})\), concave up on \((3 - 3\sqrt{2},3 + 3\sqrt{2})\cup(3 + 3\sqrt{2},\infty)\)