What is the maximum or minimum value of the function? What is the range? y = –2x^2 + 32x –12

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What is the maximum or minimum value of the function? What is the range? y = –2x^2 + 32x –12

Mathematics
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find f'(x) and equate it to 0 solve for x
I don't understand that ?
oh - i've assumed you know some calculus

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Other answers:

Would it be like y=-2(0)^2+32(0)-12 ?
no what you have there is the intercept on y-axis
you need to change it to the vertex form y = –2x^2 + 32x –12 = -2(x^2 - 16x + 6) = -2( x - 8)^2 -64 + 6) = -2[(x-8)^2 - 58)] so the x coordinate at maximum = 8 a negative coefficient before x^3 means there is a maximum value
oh , but it says the maximum and range is only 116 . its either 116 or -116 ?
at maximum the function has a value -2*-58 = 116
ok , i see that . . .
the range is f(x)<= 116
less than or greater than ?
|dw:1350681832911:dw|
below the graph less or equal to 116
Thats what i thought it was (: . Thank you soo much for helping me ! (;
yw

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