SOLUTION: What is the graph of the rational function y=x^2-4x+3/x^2-9

Hi, there--

THE PROBLEM:
What is the graph of the rational function 

A SOLUTION:
The graph of a rational function often has a restricted domain and asymptotes. 
Let's check for those.

DOMAIN:
Because we are dividing one polynomial by another, we need to make sure we do not divide 
by zero. Set the denominator equal to zero, and solve for x.



Factor the left-hand side (difference of two squares).


Apply the Zero-Product Property which basically says, "you are multiplying to factors together 
and the answer is zero. This means that at least one of those factors must equal zero too."

Either x + 3 = 0 or x - 3 = 0.

Solve for x.
x = -3 or x = 3

In other words, when x = -3 or when x = 3, the denominator will equal 0, and we will be 
dividing by zero. We need to restrict the domain to keep his from happening.

Therefore, our domain is all real numbers, such that x is not equal to 3 or -3.

ASYMPTOTES:
Most often, you will have vertical asymptotes at the zeroes of the denominator (x=-3 and 
x=3 in your problem.) 

We want to make sure that the numerator and denominator do not have common factors, though.

Factor the numerator: 
Factor the denominator: 

Notice that (x-3) is a common factor in both the numerator and denominator. 
When this occurs we will not have a vertical asymptote at x=3, but we need to check the 
function at that point. We have



In this case, the graph will have a hole in the smooth curve right at x=3 because 0/0 is not a 
real number. 

(NOTE: If you graph the function on a graphing calculator, you will not see this hole 
unless you REALLY zoom in. If you check the table of values, you will see ERROR for the 
y-value associated with x=3.)

When the degree of the polynomial in the numerator matches that of the polynomial in the 
denominator, we also have horizontal, not slant, asymptotes. To find these, divide the 
coefficient of the highest degree term in the numerator by the coefficient of the highest 
degree term in the denominator. In your problem, we have y=1/1 or y=1 as the horizontal 
asymptote. 

X- AND Y-INTERCEPTS
By solving for x = 0, you can find the y-intercept.




The graph crosses the y-axis at  (0, -1/3).

By solving for y=0, you can find the x-inercepts. In this case y=0 if the numerator equals 0. Recall that we factored the numerator earlier.



Solving for x gives x=1 or x=3. The graph crosses the the x-axis at (1,0). (At x=3, we have the hole in the graph.)

Putting this all together, we have a graph with two branches: to the left of the vertical 
asymptote the graph resides above y=1. To the right of the vertical asymptote, the graph 
resides below y=1. Remember to mark the hole with at open circle and to indicate the 
intercepts.


Here is a (long) link to the graph of this function. You will need to cut and past it into your 
web browser:

http://graphsketch.com/?eqn1_color=1&eqn1_eqn=(x%5E2-4x%2B3)%2F(x%5E2-9)&eqn2_color=2&eqn2_eqn=&eqn3_color=3&eqn3_eqn=&eqn4_color=4&eqn4_eqn=&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=-17&x_max=17&y_min=-10.5&y_max=10.5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525

Hope this helps! Feel free to email if you have questions about the explanation.
Mrs. Figgy
math.in.the.vortex@gmail.com