# Foreach of the following functions use the quadratic formula to findthe zeros of f. Then, find the maximum or minimum value of f(x). (a)f(x)=x^2 -8x (b)f(x)=-2x^2-9x+5

Foreach of the following functions use the quadratic formula to find the zeros of f. Then, find the maximum or minimum value of f(x).
(a)$f\left(x\right)={x}^{2}-8x$
(b)$f\left(x\right)=-2{x}^{2}-9x+5$
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Selden1f
a. $f\left(x\right)={x}^{2}-8x$
${x}^{2}-8x=0$
x(x-8) = 0
so, the zeros are at 0 and 8.
min val = 0
max val = 8
Since a > 0, it has a minimum. The minimum value of f(x) is-16 when x is 4. The maximum and minimum of the parabola is thevertex.So, this is determined by finding the vertex x = -b/2a, thensubstituting x to find the value of f(x)
b) $f\left(x\right)=-2{x}^{2}-9x+5$
$-\left(2{x}^{2}+9x-5\right)=0$
(2x - 1) (x + 5) = 0
So the zeros are at x = 1/2 and x=-5
Since a< 0, then it will have a maximum. Using the vertexformula again, we find the max is 121/8 when x is -9/4.