How do you graph #f(x)=(x^2-2x)/(x^2-4)# using holes, vertical and horizontal asymptotes, x and y intercepts?

The horizontal asymptote will occur at the ratio between the highest power in the numerator and in the denominator (only if the powers are equal).

As for intercepts, the graph will pass through the origin, and the origin will serve as the x and y intercept.

Here is the graph:

graph{(x^2 - 2x)/(x^2 - 4) [-10, 10, -5, 5]}

Hopefully this helps!

To graph the function f(x) = (x^2 - 2x)/(x^2 - 4), we can analyze its holes, vertical and horizontal asymptotes, as well as the x and y intercepts.

1. Holes: To find the holes, we need to determine the values of x that make the denominator (x^2 - 4) equal to zero. Solving x^2 - 4 = 0, we get x = ±2. Therefore, there are holes at x = 2 and x = -2.

2. Vertical asymptotes: Vertical asymptotes occur when the denominator of a rational function equals zero, but the numerator does not. In this case, the denominator (x^2 - 4) equals zero at x = ±2. Thus, there are vertical asymptotes at x = 2 and x = -2.

3. Horizontal asymptote: To find the horizontal asymptote, we compare the degrees of the numerator and denominator. Both have a degree of 2, so we divide the leading coefficients. In this case, the leading coefficients are both 1. Therefore, the horizontal asymptote is y = 1.

4. x-intercepts: To find the x-intercepts, we set the numerator (x^2 - 2x) equal to zero and solve for x. Factoring x(x - 2) = 0, we find x = 0 and x = 2. Thus, the x-intercepts are at x = 0 and x = 2.

5. y-intercept: To find the y-intercept, we substitute x = 0 into the function. f(0) = (0^2 - 2(0))/(0^2 - 4) = 0/(-4) = 0. Therefore, the y-intercept is at y = 0.

By considering these aspects, we can graph the function f(x) = (x^2 - 2x)/(x^2 - 4) accordingly.