If $p (x)$ $p(x)$ has a minimum value of $(- 8)$ $(-8)$ and x-intercepts at $x = - 4$ $x=-4$ and $x = - \frac{2}{3}$ $x=-2/3$ , for what values of $x$ $x$ is $p (x) \leq 0$ $p(x) <=0$ ?

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If $p (x)$ $p(x)$ has a minimum value of $(- 8)$ $(-8)$ and x-intercepts at $x = - 4$ $x=-4$ and $x = - \frac{2}{3}$ $x=-2/3$ , for what values of $x$ $x$ is $p (x) \leq 0$ $p(x) <=0$ ?

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Answer 1 · 2016-11-11T21:29:12

$x \in [- 4, - \frac{2}{3}]$ $x in [-4,-2/3]$

Explanation:

I made the assumption that you intended the second intercept to be $(- \frac{2}{3}, 0)$ $(-2/3,0)$
also, since this was asked under Quadratic Equations, I have assumed that $p (x)$ $p(x)$ is a quadratic equation.

If these assumptions are true then
since $p (x)$ $p(x)$ has a minimum value of $p (x) = - 8$ $p(x)=-8$
with intercepts (where $p (x) = 0$ $p(x)=0$ ) of $x = - 4$ $x=-4$ and $x = - \frac{2}{3}$ $x=-2/3$
then
$XXX p (x) \leq 0$ $color(white)("XXX")p(x) <=0$ when $x$ $x$ is between $(- 4)$ $(-4)$ and $(- \frac{2}{3})$ $(-2/3)$ inclusive.
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[Ask as a separate question if you need the derivation of the exact formula for $p (x)$ $p(x)$ ].

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If $p (x)$ $p(x)$ has a minimum value of $(- 8)$ $(-8)$ and x-intercepts at $x = - 4$ $x=-4$ and $x = - \frac{2}{3}$ $x=-2/3$ , for what values of $x$ $x$ is $p (x) \leq 0$ $p(x) <=0$ ?

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If p(x)p(x)p(x) has a minimum value of (-8)(−8)(-8) and x-intercepts at x=-4x=−4x=-4 and x=-2/3x=−23x=-2/3, for what values of xxx is p(x) <=0p(x)≤0p(x) <=0?

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If $p (x)$ $p(x)$ has a minimum value of $(- 8)$ $(-8)$ and x-intercepts at $x = - 4$ $x=-4$ and $x = - \frac{2}{3}$ $x=-2/3$ , for what values of $x$ $x$ is $p (x) \leq 0$ $p(x) <=0$ ?