andreascott
What is the maximum or minimum value of the function? What is the range?
y = –2x^2 + 32x –12
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cwrw238
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find f'(x) and equate it to 0
solve for x
andreascott
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I don't understand that ?
cwrw238
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oh - i've assumed you know some calculus
andreascott
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Would it be like y=-2(0)^2+32(0)-12 ?
cwrw238
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no what you have there is the intercept on y-axis
cwrw238
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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
andreascott
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oh , but it says the maximum and range is only 116 . its either 116 or -116 ?
cwrw238
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at maximum the function has a value -2*-58 = 116
andreascott
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ok , i see that . . .
cwrw238
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the range is f(x)<= 116
andreascott
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less than or greater than ?
cwrw238
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|dw:1350681832911:dw|
cwrw238
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below the graph less or equal to 116
andreascott
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Thats what i thought it was (: . Thank you soo much for helping me ! (;
cwrw238
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yw