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

Explanation:

Step1: Find the domain

Set 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 $3x-4=0$, so $x=\frac{4}{3}$.

Step3: Find the y - intercept

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

Step4: Find the vertical asymptotes

Since the denominator is zero at $x=-1$ and $x = 1$, the vertical asymptotes are $x=-1$ and $x = 1$.

Step5: Find the horizontal asymptote

Degree of numerator is 1 and degree of denominator is 2. As $x\to\pm\infty$, $y\to0$. 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 $(-\infty,-1)$ take $x=-2$, $y=\frac{3\times(-2)-4}{(-2)^{2}-1}=\frac{-6 - 4}{4 - 1}=-\frac{10}{3}<0$. In $(-1,1)$ take $x = 0$, $y = 4>0$. In $(1,\infty)$ take $x = 2$, $y=\frac{3\times2-4}{2^{2}-1}=\frac{2}{3}>0$.

Answer:

Based on the above - mentioned characteristics (x - intercept at $x=\frac{4}{3}$, y - intercept at $y = 4$, vertical asymptotes at $x=-1$ and $x = 1$, horizontal asymptote at $y = 0$ and the sign analysis), we can choose the correct graph. Without seeing the actual graphs in detail, but by following these steps, you can match the graph to the function's properties. If you provide the visual details of the graphs A, B, C, D, we can further determine the exact answer.