Substitute #y# for #f(x)#.
#y=(2x^2+5x-12)/(x+4)#
Factor the numerator using the #a*c# method.
#2x^2+5x-12#
#ax^2+bx+c#
#a=2;# #b=5;# #c=-12#
#a*c=2*-12=-24#
Find two numbers that when multiplied equal #-24# and when added equal #5#.
The numbers #-3# and #8# fit the criteria.
Rewrite #5x# as #-3x# and #8x#.
#2x^2-3x+8x-12#
Group and factor.
#(2x^2-3x)+(8x-12)# =
#x(2x-3)+4(2x-3)# =
#(x+4)(2x-3)#
Rewrite the numerator as #(x+4)(2x-3)#.
#y=((x+4)(2x-3))/(x+4)#
Cancel #(x+4)#.
#y=(cancel(x+4)(2x-3))/cancel(x+4)# =
#y=2x-3#
Make a table of #x# and #y#. Plot the points, and draw a line through the points.
Table of #x# and #y# values.
#x=-2;# #y=-7#
#x=0;# #y=-3#
#x=2;# #y=1#
graph{y=2x-3 [-11.3, 11.2, -7.56, 3.69]}
