Find the horizontal asymptotes. Since the degree of numerator and denominator are the same, we simply divide the leading coefficients.
y = 4
Now we examine r(x) to see if we get positive and negative values.
r(x) = 4x2 / (x - 3)(x + 1)
with vertical asymptotes x=-1 and x=3
The numerator will always be positive. But what about the denominator?
r(x) is positive in the interval (-∞, -1)∪(3, ∞)
r(x) is negative in the interval (-1, 3)
So we know the function goes below and above the x-axis.
What you can do next is find the local maximum and local minimum. By taking the first derivative of r(x) and setting it equal to zero, solving for x, you can find the location of the maximum and minimum.
f'(x) = 0
[8x(x2 - 2x - 3) - 4x2(2x - 2)] / (x2 - 2x - 3)2 = 0
(-8x2 - 24x) / (x2 - 2x - 3)2 = 0
Setting numerator equal to zero,
-8x(x + 3) = 0
x = 3 and x = 0
We see that r(-3) is a minimum and r(0) is a maximum using 1st derivative test.
The minimum, maximum, and horizontal asymptote are clues to find your range. Find the values of the max and min and see how they relate to the horizontal asymptote. This is because there are some restrictions on the range.