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

Question

Isaiah Anthony · Accepted Answer

See below.

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!

Hazel Anglin · Answer

See graph

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]}

Genesis Allen · Answer

undefined To graph the function f(x) = (x^2-5x+6)/(x^2-4x+3), we can analyze its holes, vertical and horizontal asymptotes, as well as the x and y intercepts.

First, let's find the x-intercepts by setting f(x) equal to zero and solving for x:
(x^2-5x+6)/(x^2-4x+3) = 0
x^2-5x+6 = 0
(x-2)(x-3) = 0
x = 2 or x = 3

Therefore, the x-intercepts are (2, 0) and (3, 0).

Next, let's find the y-intercept by evaluating f(x) when x = 0:
f(0) = (0^2-5(0)+6)/(0^2-4(0)+3)
f(0) = 6/3
f(0) = 2

Therefore, the y-intercept is (0, 2).

To determine the holes, we need to factor the numerator and denominator:
x^2-5x+6 = (x-2)(x-3)
x^2-4x+3 = (x-1)(x-3)

We can see that (x-3) is a common factor in both the numerator and denominator. Therefore, there is a hole at x = 3.

To find the vertical asymptotes, we need to determine the values of x that make the denominator equal to zero:
x^2-4x+3 = 0
(x-1)(x-3) = 0
x = 1 or x = 3

Therefore, there are vertical asymptotes at x = 1 and x = 3.

To find the horizontal asymptote, we need to compare the degrees of the numerator and denominator. Since both have a degree of 2, we divide the leading coefficients:
Leading coefficient of numerator: 1
Leading coefficient of denominator: 1

Therefore, the horizontal asymptote is y = 1.

To summarize:
- x-intercepts: (2, 0) and (3, 0)
- y-intercept: (0, 2)
- Hole: x = 3
- Vertical asymptotes: x = 1 and x = 3
- Horizontal asymptote: y = 1