How do you graph #f(x)=(x^2-5x+6)/(x^2-4x+3)# using holes, vertical and horizontal asymptotes, x and y intercepts?

2 Answers

See below.

Explanation:

It would be very hard to graph this equation correctly as without factoring both the numerator and denominator it would be hard to account for all of the information.

#x^2 - 5x + 6#
#x^2 - 2x - 3x - 6#
#(x-2)(x-3)#

Then, following the same process for the bottom, you get

#(x-1)(x-3)#

Based on this, you can rewrite f(x) as:

#f(x) = [(x-2)(x-3)] / [(x-1)(x-3)]#

Now you are ready to graph with all of the criteria. Since #(x-3)# cancels out on both top and bottom, you know that there must be a hole at #x = 3#. To find the #y#-coordinate of the hole, simply substitute #x = 3# into the factored equation, and the result is the #y# - coordinate. Mark the hole's coordinates with a small, open circle.

Vertical asympotes can be found in the denominator, just set #(x-1)# equal to zero. Then, solving for #x#, you get #x = 1#. The vertical asymptote of #f(x)# is #x = 1#.

For the horizontal asymptote, in this case, you would take the ratio of the leading coefficients of the numerator and denominator. Since this is #1#, then the horizontal asymptote of #f(x)# is #y = 1#.

Zeros are found in the top. You set #(x-2)# equal to zero, and the result is #x = 2#. Therfore, there must be a "zero" or x-intercept at #(2, 0)#.

You know the #y# - intercept is #(0, 2)# by substituting #x = 0# into the equation.

That's all. Hope that helps!

Apr 24, 2017

See graph

Explanation:

In addition to to the answer below, here is what the graph would look like. Keep in mind however, that this graph show the "hole" at #(3, 1/2)# that would be in the graph.

graph{(x^2-5x+6)/(x^2-4x+3) [-10, 10, -5, 5]}