make a complete graph of the following function. a graphing utility is useful in locating intercepts, local…

Explanation:

Step1: Find the domain

The denominator $x^{2}-1=(x - 1)(x + 1)\neq0$, so $x\neq\pm1$. The domain is $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.

Step2: Find the x - intercept

Set $y = 0$, then $\frac{3x-4}{x^{2}-1}=0$, so $3x - 4=0$ and $x=\frac{4}{3}$.

Step3: Find the y - intercept

Set $x = 0$, then $y=\frac{3\times0 - 4}{0^{2}-1}=4$.

Step4: Find the vertical asymptotes

Since $x^{2}-1 = 0$ when $x=\pm1$, the vertical asymptotes are $x = 1$ and $x=-1$.

Step5: Find the horizontal asymptote

As $x\to\pm\infty$, we use the fact that for $y=\frac{3x-4}{x^{2}-1}$, since the degree of the numerator is 1 and the degree of the denominator is 2, $y\to0$ as $x\to\pm\infty$, so the horizontal asymptote is $y = 0$.

Step6: Analyze the sign of the function

We can use test - points in the intervals $(-\infty,-1),(-1,1),(1,\infty)$. For example, in $(-\infty,-1)$ let $x=-2$, then $y=\frac{3\times(-2)-4}{(-2)^{2}-1}=\frac{-6 - 4}{4 - 1}=-\frac{10}{3}<0$. In $(-1,1)$ let $x = 0$, $y = 4>0$. In $(1,\infty)$ let $x = 2$, then $y=\frac{3\times2-4}{2^{2}-1}=\frac{6 - 4}{4 - 1}=\frac{2}{3}>0$.

Based on the above - mentioned characteristics (x - intercept at $x=\frac{4}{3}$, y - intercept at $y = 4$, vertical asymptotes $x=\pm1$, horizontal asymptote $y = 0$, and the sign of the function in different intervals), we can determine the graph.

Answer:

(Without seeing the actual options clearly, the above steps help in identifying the correct graph. You need to match the graph based on the x - intercept, y - intercept, asymptotes and sign analysis. If you provide the details of the graphs in text form, a more definite answer can be given.)