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

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Answer 1 · 2018-07-16T10:45:25

$x = - 2, 2, 3$ $x=-2,2,3$

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

$2 (x^{2} - 4) (x - 3) = 0$ $2(x^2-4)(x-3)=0$
and we get

$x^{2} - 4 = 0$ $x^2-4=0$ so $x_{1} = 2$ $x_1=2$ or $x_{2} = - 2$ $x_2=-2$

or

$x - 3 = 0$ $x-3=0$ so $x_{3} = 3$ $x_3=3$
and the y -intercept is $p (0) = 2 \cdot (- 4) \cdot (- 3) = 24$ $p(0)=2*(-4)*(-3)=24$

My corrections.

Answer 2 · 2018-07-16T10:48:05

$y-intercept = 24$ $"y-intercept "=24$

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

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

What is the yy-intercept of p(x)=2(x2−4)(x−3)p(x)=2(x^2-4)(x-3)?