What are the #x#-intercepts of the function #f(x) = -4x^2-16x-12#?

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Answer 1 · 2017-05-16T01:30:19

The x-intercepts are #(-1,0)# and #(-3,0)#.

The x-intercepts are the solutions to #x# when #y=0#.

Substitute #0# for #f(x)#.

#-4x^2-16x-12=0#

Factor out the common term #-4#.

#-4(x^2+4x+3)=0#

Factor #x^2+4x+3#.

Find two numbers that when added equal #4# and when multiplied equal #3#. The numbers #1# and #3# meet the requirements.

Rewrite the equation as two binomials.

#-4(x+1)(x+3)=0#

Divide both sides by #-4#.

#(-4(x+1)(x+3))/-4=0/(-4)#

#(color(red)cancel(color(black)(-4))(x+1)(x+3))/(color(red)cancel(color(black)(-4)))=0/(-4) #

#(x+1)(x+3)=0#

Solutions for #x#.

#x=-1,-3#

x-intercepts

#(-1,0)#
#(-3,0)#
graph{-4(x^2+4x+3)=y [-11.16, 11.34, -6.975, 4.275]}