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Mathematics 8 Online
OpenStudy (anonymous):

What is the relative maximum and minimum of the function? f(x) = 2x2 + 28x - 8

OpenStudy (anonymous):

A. Minimum Value: -7 Range y > -7 B. Minimum Value: 7 Range y > 7 C. Minimum Value: -106 Range y > -106 D. Minimum Value: -106 Range y > -7

OpenStudy (dan815):

find the vertex of this parabola

OpenStudy (dan815):

its a parabola facing upwards so, you can tell the minimum y value from the vertex

OpenStudy (anonymous):

I don't know how to

OpenStudy (dan815):

and the Range, meaning all the other y values. will include every other y number greater than that min value

OpenStudy (dan815):

have you learnt about derivatives yet?

OpenStudy (anonymous):

yes but I really didn't understand

OpenStudy (anonymous):

I had thought B but it was wrong

OpenStudy (dan815):

for a parabola, the maximum or minimum happens when the derivative =0, because the tangent line there must be horizontal|dw:1403083206816:dw|

OpenStudy (anonymous):

ok

OpenStudy (dan815):

f'(x) = ?

OpenStudy (anonymous):

2x^2 +28x -8

OpenStudy (dan815):

f ' (x) = take the derivative of f(x)

OpenStudy (dan815):

use this formula for now it comes straight from differentiation -b/2a will give you the x value where the vertex is -28/4 = -7 y=2*(-7)^2+28*-7 -8 = 98 -140-56-8 = -106

OpenStudy (dan815):

therefre min value = -106, and the range of y is every value more or equal to that

OpenStudy (anonymous):

so C??

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