Questions: Make a sign diagram for the derivative of the rational function. f(x) = (2x+6)/(x-3) f'(x)<0 v f'(x)=0 x f'(x)<0 v x= Find all relative extreme points. (If an answer does not exist, enter DNE.) relative max (x, y)= relative min (x, y)= Find all asymptotes. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) vertical asymptote(s) x= horizontal asymptote(s) y=

Make a sign diagram for the derivative of the rational function.
f(x) = (2x+6)/(x-3)

f'(x)<0 v f'(x)=0 x f'(x)<0 v
x=

Find all relative extreme points. (If an answer does not exist, enter DNE.)
relative max (x, y)=
relative min (x, y)=
Find all asymptotes. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
vertical asymptote(s) x=
horizontal asymptote(s) y=

Transcript text: Make a sign diagram for the derivative of the rational function. \[ f(x)=\frac{2 x+6}{x-3} \] \begin{tabular}{|l|c|c|} \hline$f^{\prime}(x)<0 \quad v$ & $f^{\prime}(x)=0 \times$ & $f^{\prime}(x)<0$ v \\ \hline & $x=\square$ & \\ \hline \end{tabular} Find all relative extreme points. (If an answer does not exist, enter DNE.) relative max $(x, y)=(\square)$ relative $\min \quad(x, y)=$ $\square$ Find all asymptotes. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) vertical asymptote(s) $\quad x=$ $\square$ horizontal asymptote(s) $\quad y=$ $\square$

Solution

Solution Steps

Step 1: Find the Derivative of the Function

To find the derivative of the function $ f(x) = \frac{2x + 6}{x - 3} $, we use the quotient rule: \[ f'(x) = \frac{(x-3)(2) - (2x+6)(1)}{(x-3)^2} = \frac{2x - 6 - 2x - 6}{(x-3)^2} = \frac{-12}{(x-3)^2} \]

Step 2: Determine the Sign of the Derivative

The derivative $ f'(x) = \frac{-12}{(x-3)^2} $ is always negative because the numerator is negative and the denominator is always positive for $ x \neq 3 $.

Step 3: Find Relative Extreme Points

Since $ f'(x) $ is never zero and does not change sign, there are no relative extreme points.

Final Answer

Relative max: DNE
Relative min: DNE

Step 4: Find Asymptotes

Vertical asymptote: Set the denominator equal to zero, $ x - 3 = 0 $, so $ x = 3 $.
Horizontal asymptote: As $ x \to \infty $, the degree of the numerator and denominator are the same, so the horizontal asymptote is $ y = \frac{2}{1} = 2 $.

Final Answer

Vertical asymptote(s): $ x = 3 $
Horizontal asymptote(s): $ y = 2 $

{"axisType": 3, "coordSystem": {"xmin": -10, "xmax": 10, "ymin": -10, "ymax": 10}, "commands": ["y = (2x + 6)/(x - 3)"], "latex_expressions": ["$y = \\frac{2x + 6}{x - 3}$"]}

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