What is the #y#-intercept of #p(x)=2(x^2-4)(x-3)#?

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Answer 1 · 2018-07-16T10:45:25

#x=-2,2,3#

For the y-intercept must be
#p(x)=0#
this

#2(x^2-4)(x-3)=0#
and we get

#x^2-4=0# so #x_1=2# or #x_2=-2#

or

#x-3=0# so #x_3=3#
and the y -intercept is #p(0)=2*(-4)*(-3)=24#

My corrections.

Answer 2 · 2018-07-16T10:48:05

#"y-intercept "=24#

#"set x = 0 for y-intercept"#

#p(0)=2(0-4)(0-3)=2(-4)(-3)=24#