How do you graph #f(x)=(x+2)(x-1)/(x-3)#?

1 Answer
Jul 24, 2018

See answer below

Explanation:

Given: #f(x) = (x+2)((x-1))/(x-3)#

#f(x) = (x+2)((x-1))/(x-3) = (x+2)/1((x-1))/(x-3) = ((x+2)(x-1))/(x-3)#

This type of equation is called a rational (fraction) function.

It has the form: #f(x) = (N(x))/(D(x)) = (a_nx^n + ...)/(b_m x^m + ...)#,

where #N(x))# is the numerator and #D(x)# is the denominator,

#n# = the degree of #N(x)# and #m# = the degree of #(D(x))#

and #a_n# is the leading coefficient of the #N(x)# and

#b_m# is the leading coefficient of the #D(x)#

Step 1 factor : The given function is already factored.

Step 2, cancel any factors that are both in #(N(x))# and #D(x))# (determines holes):

The given function has no holes #" "=> " no factors that cancel"#

Step 3, find vertical asymptotes: #D(x) = 0#

vertical asymptote at #x = 3#

Step 4, find horizontal asymptotes:
Compare the degrees:

If #n < m# the horizontal asymptote is #y = 0#

If #n = m# the horizontal asymptote is #y = a_n/b_m#

If #n > m# there are no horizontal asymptotes

In the given equation: #n = 2; m =1 #

There is no horizontal asymptotes

Step 5, find slant or oblique asymptote when #n = m+1#:

#f(x) = ((x+2)(x-1))/(x-3) = (x^2 +x - 2)/(x-3)#

Use Long division:

#" "ul(" "color(red)(x + 4) +10/(x-3))#
#x-3 | x^2 + x - 2#
#" "ul(x^2 - 3x)#
#" "4x - 2#
#" "ul(4x - 12)#
#" "10#

slant or oblique asymptote: #color(red)(y = x+4)#

Step 6, find x-intercept(s) : #N(x) = 0#

#(x + 2)(x-1) = 0#
# " "x + 2 = 0 " "=> x"-intercept" (-2, 0)#
# " "x -1 = 0 " "=> x"-intercept" (1, 0)#

Step 7, find y-intercept(s): #x = 0#

#f(0) = (0+2)((0-1))/(0-3) = 2(-1)/-3 = 2/3#

#y#-intercept: (0, 2/3)#

Graph of #f(x) = (x+2)((x-1))/(x-3)#:

graph{(x+2)((x-1))/(x-3) [-50, 50, -50, 50]}