Christopher G. answered • 08/13/13

Math Tutor - Algebra, Trig, Calculus, SAT/ACT Math

When graphing a rational function, find the x-intercepts, y-intercepts, vertical asymptotes, and horizontal asymptotes.

**x-intercepts: **x-values when y = 0 (hint: y = 0 only when the numerator = 0)

0 = x + 3 -> x = -3, or (-3, 0)

**y-intercepts:** y-values when x = 0

y = (0 + 3)/(0 - 3) = 3/-3 -> y = -1, or (0, -1)

**VA:** x-values when the function is undefined (denominator = 0)

0 = x - 3 -> x = 3

**HA:** there are three rules for horizontal asymptotes:

(1) if the degree of the numerator < the degree of the denominator, then the HA is y = 0

(2) if the degree of the numerator = the degree of the denominator, then the HA is y = c, where c is the ratio of the leading coefficients in the numerator and denominator

(3) if the degree of the numerator > the degree of the denominator, then there is no HA

In this case, both the numerator and denominator have a degree of 1, so the HA is at some constant c; the ratio of the leading coefficients is 1/1 = 1, therefore the HA is y = 1.

A function may never cross a VA, but it can cross a HA, so you need to check that. Set the function equal to the HA value.

1 = (x + 3)/(x - 3) -> x - 3 = x + 3 -> 0 ≠ 6 -> there are no x-values that cross the HA

Now you need to test where the graph lies in relation to your x-intercept, x = -3, and vertical asymptote, x = 3. Make the following intervals: (-∞, -3), (-3, 3), and (3, ∞).

In the first interval, (-∞, -3), the function is positive. Test a value in this interval, say x = -10, to see why. Since we know the function doesn't cross the HA, it must be either entirely above it or entirely between the HA and the x-axis to stay positive. We know the graph must cross the x-axis at x = 3, so therefore the function lies between the HA and the x-axis in this interval.

In the second interval, (-3, 3), the function is negative. Test x = 0 to see why. We know the function continues to -∞ as it approaches the VA.

In the third interval, (3, ∞), the function is positive. Test x = 5 to see why. Again, we know the function doesn't cross the HA, so it must be either entirely above it or between it and the x-axis. Since the function is positive, it must be approaching +∞ at the VA, so the function must exist entirely above the HA in this interval.

Hope this helps!