ExtremeFind the extreme of the function y=x^2-8x+16.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may also use derivative of the function to evaluate its extreme value, such that:

`f'(x) = (x^2 - 8x + 16)' => f'(x) = (x^2)' - (8x)' + 16'`

`f'(x) = 2x - 8 + 0 => f'(x) = 2x - 8`

You need to solve for x the equation `f'(x) = 0` , such that:

`{(f'(x) = 0),(f'(x) = 2x - 8):} => 2x - 8 = 0 => 2x = 8 => x = 8/2 => x = 4`

Hence, the function reaches its extreme at `x = 4` , thus, evaluating the y coordinate of extreme point yields:

`f(4) = 4^2 - 8*4+ 16 => f(4) = 32 - 32 => f(4) = 0`

Hence, evaluating the extreme point of the function yields that the function reaches its extreme point at `x = 4, y = 0.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To extreme of a function, whose expression is a quadratic, is the vertex of the parable f(x) = y.

We know that the coordinates of the parabola vertex are:

V(-b/2a;-delta/4a), where a,b,c are the coefficients of the  function and delta=b^2 -4*a*c.

y=f(x)=x^2 - 8x + 16

We'll identify the coefficients:

a=1, 2a=2, 4a=4

b=-8, c=16

delta=(-8)^2 -4*1*16

delta =64 - 64

delta = 0

V(-b/2a;-delta/4a)=V(-(-8)/2;-(0)/4)

V(4;0)

Since the coefficient of x^2 is positive, the extreme point is a minimum point.

Because the x coordinate is positive and y coordinate is 0, the vertex is located on the right side of x axis: V(4;0).

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