Tools≻Load Package: Student Calculus 1

Control-drag $f\left(x\right)\=\dots$ 

Context Panel: Assign Function

The graph of $f$ appears to have $h\=0$ as a horizontal asymptote for all values of $r$.

For values of $r$ less than some threshold value near $r\=9$, the graph of $f$ appears to have two vertical asymptotes, and a single relative maximum that is negative.

For values of $r$ greater than this threshold, the graph of $f$ appears to have no vertical asymptotes, and a single relative maximum that is positive, and is also an absolute maximum.

At the threshold value of $r$, it appears that the two vertical asymptotes in the graph of $f$ merge into a single vertical asymptote, and there are no extrema.

Type $f\left(x\right)$.

Context Panel: Evaluate and Display Inline

Context Panel: Complete Square≻$x$

There are two vertical asymptotes; equations are $x\&equals;x_1$ and $x\&equals;x_2$, with $x_1<x_2$.

For $x<x_1$, $f$ is increasing and concave upward because $f\prime$and $f″$ are positive.

For $x\&gt;x_2$, $f$ is decreasing but concave upward because $f\prime$is negative and $f″$ is positive.

For $x\in \left(x_1\&comma;x_2\right)$, $f$ increases to a local maximum (at $x\&equals;-3$), then decreases;
(The first derivative goes from positive to negative.)

For $x\in \left(x_1\&comma;x_2\right)$, $f$ is concave downward because $f″$ is strictly negative.

On any finite domain that included the two vertical asymptotes, the endpoints would be local minima.

The single vertical asymptote has the equation $x\&equals;-3$.

For $x<-3$, $f$ is increasing and concave upward because both $f\prime$ and $f″$ are positive.

For $x\&gt;-3$, $f$ is decreasing and concave upward because $f\prime$ is negative while $f″$ is positive.

On any finite domain that included the vertical asymptote, the endpoints would be local minima.

For all $r\&gt;9$, there is a local maximum at $x\&equals;-3$, and two inflection points symmetrically placed, one on either side of $x\&equals;-3$.

For $x<-3$, $f$ is increasing because $f\prime$ is positive. However, $f$ changes concavity from upward to downward because $f″$ changes from positive to negative.

For $x\&gt;-3$, $f$ is decreasing because $f\prime$ is negative. However, $f$ changes concavity from downward to upward because $f″$ changes from negative to positive.

On any finite domain that included $x\&equals;-3$, the endpoints would be local minima.

Write $f\prime \left(x\right)$
Context Panel: Evaluate and Display Inline

Context Panel: Complete Square≻$x$

Context Panel: Solve≻Obtain Solutions for≻$x$ 

Expression palette: Evaluation template
Evaluate $f\prime \left(x\right)$ for $r\&equals;9$
Context Panel: Evaluate and Display Inline

Context Panel: Factor

Write $f″\left(x\right)\&equals;0$; press the Enter key.

Context Panel: Simplify≻Simplify

Context Panel:
Solve≻Obtain Solutions for≻$x$ 

Context Panel: Conversions≻To List

Context Panel: Assign to a Name≻$p$

The graph in Figure 3.7.6(f) is generated by the FunctionChart command applied to $f$, with the value of $r$ controlled by a slider.

From Figures 3.7.6(c-e) and the Curve Analysis tutor, the data in Table 3.7.6(e) can be constructed, provided $a_{\pm }\&equals;-3 \pm \sqrt{9-r}$, and $p_{\pm }\&equals;-3 \pm \sqrt{\left(r-9\right)\&sol;3}$ are calculated analytically as the locations of the vertical asymptotes, and inflection points, respectively. Table 3.7.6(e) assumes that the domain of $f$ is as large as possible. Note that $f$ has no zeros or local minima.

To launch the Curve Analysis tutor for $f$:
Enter a value of $r$: ;
then press:  tutor.