How do you graph # f(x)= (2x^2+5x-12)/(x+4)#?

1 Answer
Jul 1, 2015

Factor the numerator using the #a*c# method. Cancel common terms in the numerator and denominator. Make a table of #x# and #y# values. Plot the points, and draw a straight line through the points.

Explanation:

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]}