Question

Sketch the graph of the function. Label the intercepts, relative extrema, points of inflection, and asymptotes. Then state the domain of the function. \(y=\frac{x^{2}+1}{x^{2}-9}\)

Short answer

The graph of the function \(y=\frac{x^{2}+1}{x^{2}-9}\) has a y-intercept at -1/9, vertical asymptotes at x=-3 and x=3, a horizontal asymptote at y=1, no relative extrema or points of inflection, and a domain of \(x \in (-\infty, -3) \cup (-3, 3) \cup (3, +\infty)\).

Step-by-step solution

Determine the x and y intercepts

Set \(y=0\) to find the x-intercepts. There are none because the numerator is \(x^{2}+1\), which is always positive. To find the y-intercept, set \(x=0\) and you get \(y=\frac{1}{-9}\) or \(y=-\frac{1}{9}\). So the y-intercept is -1/9.

Find the vertical and horizontal asymptotes

To find the vertical asymptotes, set the denominator \(x^{2}-9\) equal to zero and solve for x. You get x=-3 and x=3. The horizontal asymptote can be found by taking the limit as x approaches infinity and negative infinity. The coefficients of the highest degree terms in the numerator and denominator are the same, therefore the horizontal asymptote is y=1.

Determine the relative extrema and points of inflection

Take the derivative of the function and set it equal to 0 to find the critical points, then use the second derivative test to determine whether these are maxima, minima, or points of inflection. For this function, the derivative is complex and no real critical points are found. Therefore, this function has no relative extrema or points of inflection.

State the domain of the function

The domain of the function is all real numbers except for 3 and -3, which are the vertical asymptotes. So the domain is \(x \in (-\infty, -3) \cup (-3, 3) \cup (3, +\infty)\)