(a)$f(x)={x}^{2}-8x$

(b)$f(x)=-2{x}^{2}-9x+5$

kadejoset
2022-07-28
Answered

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)$f(x)={x}^{2}-8x$

(b)$f(x)=-2{x}^{2}-9x+5$

(a)$f(x)={x}^{2}-8x$

(b)$f(x)=-2{x}^{2}-9x+5$

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Selden1f

Answered 2022-07-29
Author has **14** answers

a. $f(x)={x}^{2}-8x$

${x}^{2}-8x=0$

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) $f(x)=-2{x}^{2}-9x+5$

$-(2{x}^{2}+9x-5)=0$

(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.

${x}^{2}-8x=0$

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) $f(x)=-2{x}^{2}-9x+5$

$-(2{x}^{2}+9x-5)=0$

(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.

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