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

1 Answer
Jun 12, 2018

Answer:

Below

Explanation:

#f(x)=(x-4)/(x^2-3x)#

#f(x)=(x-4)/(x(x-3))#

For vertical asymptotes, we look at the denominator and since the denominator cannot equal to 0, we let the denominator equal to 0 to see where the graph cannot be

Vertical Asmpytote
#x(x-3)=0#
#x=0# or #x=3#

For horizontal asymptote, since the degree of the numerator is smaller than the denominator, then the line #y=0# is the horizontal asymptote. If the degree of the numerator is greater than the denominator, we need to use long division to find the oblique asymptote and not the horizontal asmpytote.

In this case, since the degree of the numerator is greater than the denominator, then #y=0# is the horizontal asymptote.

For x-intercepts, let #y=0#

#0=(x-4)/(x^2-3x)#
#0=x-4#
#x=4# #(4,0)# is the x-intercept

For y-intercepts, let #x=0#
But there is no solution since the denominator will equal to 0.

Your final graph should look something like this:
graph{(x-4)/(x^2-3x) [-14.24, 14.24, -7.12, 7.12]}