how to graph a rational function

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!

Use the substitution x:

x=0 ...... y = (0+3)/(0-3)=3/-3=-1 ..... point x=0 y=-1

x=1 ...... y = (1+3)/(1-3)=4/-2=-2 ..... point x=1 y=-2

x=2 ...... y = (2+3)/(2-3)=5/-1=-5 ..... point x=2 y=-5

and etc.... in the opposite direction the same

Step 1 - Rewrite the left-hand side as a single "fraction" or "rational expression". Hint: the common denominator is "x"...

Step 2 - Find the a) roots and b) discontinuities of the rewritten rational expression.
    a) Since the numerator is now a quadratic expression, factor or use the quadratic formula to find its roots.
    b) The discontinuity is where ever the denominator equals 0. From my "hint" in Step 1 (since the denominator is "x") it will have a discontinuity at x = 0.

Step 3 - Plug in a number between each of the roots and discontinuities and determine just the "sign" (+ or -) of the output. This will give you a general idea of which way the graph "bends" (+ = up, - = down) and should be enough to sketch out the rest of the function.

Related sites:
http://www.coolmath.com/algebra/23-graphing-rational-functions/
http://tutorial.math.lamar.edu/Classes/Alg/GraphRationalFcns.aspx
http://patrickjmt.com/find-asymptotes-of-a-rational-function-vertical-and-obliqueslant/