Let's go through each part of this problem step-by-step.

Given: $r(x) = -\frac{2}{(3 - 2x)(x + 1)} + 2$

(a) End Behavior

The end behavior is determined by the behavior of $r(x)$ as $x \to \pm \infty$. Since $r(x)$ has a rational component with a constant numerator and quadratic terms in the denominator, the function approaches the horizontal asymptote defined by the constant $+2$ as $x \to \pm \infty$.

So, the end behavior: $\lim_{x \to \pm \infty} r(x) = 2$

(b) Domain and Vertical Asymptotes

To find the domain and vertical asymptotes, we need to identify values of $x$ that make the denominator zero. The denominator is: $(3 - 2x)(x + 1)$ Setting each factor to zero gives:

1. $3 - 2x = 0 \Rightarrow x = \frac{3}{2}$
2. $x + 1 = 0 \Rightarrow x = -1$

Thus, vertical asymptotes occur at $x = \frac{3}{2}$ and $x = -1$.

The domain of $r(x)$ is all real numbers except $x = \frac{3}{2}$ and $x = -1$.

(c) Local Behavior Near the Vertical Asymptotes

To describe the behavior near the asymptotes:

• As $x \to -1^{+}$, $r(x) \to +\infty$ (since the $(x + 1)$ term approaches zero positively).

• As $x \to -1^{-}$, $r(x) \to -\infty$ (since the $(x + 1)$ term approaches zero negatively).

• As $x \to \left( \frac{3}{2} \right)^{+}$, $r(x) \to -\infty$ (since $(3 - 2x)$ approaches zero negatively).

• As $x \to \left( \frac{3}{2} \right)^{-}$, $r(x) \to +\infty$ (since $(3 - 2x)$ approaches zero positively).

(d) Evaluate and Sketch the Graph

To evaluate $r(x)$ at $x = \frac{1}{4}$, substitute $x = \frac{1}{4}$ into the function: $r\left(\frac{1}{4}\right) = -\frac{2}{(3 - 2 \cdot \frac{1}{4})(\frac{1}{4} + 1)} + 2$ Simplifying each part:

1. $3 - 2 \cdot \frac{1}{4} = 3 - \frac{1}{2} = \frac{5}{2}$
2. $\frac{1}{4} + 1 = \frac{5}{4}$

Thus, $r\left(\frac{1}{4}\right) = -\frac{2}{\frac{5}{2} \cdot \frac{5}{4}} + 2 = -\frac{2}{\frac{25}{8}} + 2 = -\frac{2 \cdot 8}{25} + 2 = -\frac{16}{25} + 2 = \frac{50 - 16}{25} = \frac{34}{25}$

So, $r\left(\frac{1}{4}\right) = \frac{34}{25} \approx 1.36$.

For sketching, you would plot the asymptotes, the point at $x = \frac{1}{4}$ with $y \approx 1.36$, and another point as desired.

(e) Range of $r(x)$

Considering the end behavior, vertical asymptotes, and the points evaluated, $r(x)$ has values approaching infinity near the asymptotes, and approaches 2 as $x \to \pm \infty$. Thus, the range of $r(x)$ is all real values $(-\infty, \infty)$.

(f) Solve for $x$ when $r(x) = 0$

Set $r(x) = 0$: $-\frac{2}{(3 - 2x)(x + 1)} + 2 = 0$ Solving for $x$: $-\frac{2}{(3 - 2x)(x + 1)} = -2$ $\frac{2}{(3 - 2x)(x + 1)} = 2$ $(3 - 2x)(x + 1) = 1$

Expanding and solving this equation will give the values of $x$ when $r(x) = 0$.

Would you like further details on any specific part or help with graphing?

Here are 5 related questions to deepen understanding:

1. How would the end behavior change if the constant 2 in $r(x)$ was replaced by another value?
2. What happens to the function if we change the sign of the numerator in $r(x)$?
3. Can vertical asymptotes intersect with the graph of the function?
4. How does the range change if we have additional transformations applied to $r(x)$?
5. How can we determine the symmetry of a rational function like $r(x)$?

Tip: Always check for points where the function is undefined; these are typically where vertical asymptotes or holes might occur in rational functions.