Finding the value of #x# given #y# is a maximum? See picture below

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1 Answer
Douglas K.
Apr 9, 2017

When given an equation of the form #y = ax^2+bx+c#, where "a" is negative, the maximum value is the y coordinate of the vertex. The x coordinate of the vertex is:
#h=-b/(2a)#

Explanation:

Given: #y = -16x^2 + 160x - 256#

We observe that #a = -16# and #b = 160#

The x coordinate of the vertex is:

#h = -b/(2a)#

#h = -160/(2(-16))#

#h = 5 larr# this is the answer to [7.a.i]

The maximum value is the function evaluated at #x = h = 5#:

#y = -16(5)^2+160(5)-256#

#y = 144 larr# this is the answer to [7.a.ii]

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