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

$"y-intercept "=24$

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

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

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