Question

use the given information to sketch the graph of f. domain: all real x, except x = -6 and x = 6. f(-9) = -2; f(0) = 0; f(9) = 2. f(x) < 0 on (-∞, -6) and (6, ∞); f(x) > 0 on (-6, 6). f(x) < 0 on (-∞, -6) and (-6, 0); f(x) > 0 on (0, 6) and (6, ∞). vertical asymptotes: x = -6 and x = 6; horizontal asymptote: y = 0. choose the correct graph below.

Explanation:

Step1: Analyze domain and asymptotes

The domain has holes at \(x = - 6\) and \(x = 6\) (vertical asymptotes). Horizontal asymptote \(y = 0\) means as \(x\to\pm\infty\), \(y\to0\).

Step2: Analyze first - derivative

\(f^{\prime}(x)<0\) on \((-\infty,-6)\) and \((6,\infty)\) (function decreasing there). \(f^{\prime}(x)>0\) on \((-6,6)\) (function increasing there). So \(x=-6\) is a local minimum (from left - decreasing, right - increasing) and \(x = 6\) is a local maximum (from left - increasing, right - decreasing) in their respective intervals.

Step3: Analyze second - derivative

\(f^{\prime\prime}(x)<0\) on \((-\infty,-6)\) and \((-6,0)\) (concave down). \(f^{\prime\prime}(x)>0\) on \((0,6)\) and \((6,\infty)\) (concave up). So \(x = 0\) is an inflection point.

Step4: Use function values

\(f(-9)=-2\), \(f(0) = 0\), \(f(9)=2\) to plot key points.

Since you haven't provided the actual graph options, but based on the analysis:

• The graph will have two vertical asymptotes \(x=-6\) and \(x = 6\), a horizontal asymptote \(y = 0\).
• It will be decreasing on \((-\infty,-6)\) and \((6,\infty)\), increasing on \((-6,6)\).
• Concave down on \((-\infty,-6)\) and \((-6,0)\), concave up on \((0,6)\) and \((6,\infty)\) with an inflection point at \(x = 0\) and passing through \((-9,-2)\), \((0,0)\), \((9,2)\)

If we assume the options are standard, the graph that satisfies these properties (vertical asymptotes at \(x=\pm6\), horizontal asymptote \(y = 0\), increasing on \((-6,6)\), decreasing outside, concave - down/up as per second - derivative, and passing through the given points) is the correct one.

Answer:

Step1: Analyze domain and asymptotes

The domain has holes at \(x = - 6\) and \(x = 6\) (vertical asymptotes). Horizontal asymptote \(y = 0\) means as \(x\to\pm\infty\), \(y\to0\).

Step2: Analyze first - derivative

\(f^{\prime}(x)<0\) on \((-\infty,-6)\) and \((6,\infty)\) (function decreasing there). \(f^{\prime}(x)>0\) on \((-6,6)\) (function increasing there). So \(x=-6\) is a local minimum (from left - decreasing, right - increasing) and \(x = 6\) is a local maximum (from left - increasing, right - decreasing) in their respective intervals.

Step3: Analyze second - derivative

\(f^{\prime\prime}(x)<0\) on \((-\infty,-6)\) and \((-6,0)\) (concave down). \(f^{\prime\prime}(x)>0\) on \((0,6)\) and \((6,\infty)\) (concave up). So \(x = 0\) is an inflection point.

Step4: Use function values

\(f(-9)=-2\), \(f(0) = 0\), \(f(9)=2\) to plot key points.

Since you haven't provided the actual graph options, but based on the analysis:

• The graph will have two vertical asymptotes \(x=-6\) and \(x = 6\), a horizontal asymptote \(y = 0\).
• It will be decreasing on \((-\infty,-6)\) and \((6,\infty)\), increasing on \((-6,6)\).
• Concave down on \((-\infty,-6)\) and \((-6,0)\), concave up on \((0,6)\) and \((6,\infty)\) with an inflection point at \(x = 0\) and passing through \((-9,-2)\), \((0,0)\), \((9,2)\)

If we assume the options are standard, the graph that satisfies these properties (vertical asymptotes at \(x=\pm6\), horizontal asymptote \(y = 0\), increasing on \((-6,6)\), decreasing outside, concave - down/up as per second - derivative, and passing through the given points) is the correct one.