Determine the maximum point of the curve y = x^2 - 8x + 16.

Expert Answers
justaguide eNotes educator| Certified Educator

To determine the point of maxima for the curve y = x^2 - 8x + 16, we need to find the derivative of y and equate it to zero.

y = x^2 - 8x + 16

=> y' = 2x - 8

2x - 8 = 0

=> x = 8/2

=> x = 4

For x = 4, y = 4^2 - 8*4 + 16

=> y = 16 - 32 + 16 = 0

But we see that the point (4, 0) is an extreme point where the function has the minimum value.

It is not possible to determine the point where the function has the maximum value.

neela | Student

To determine the maximum point of the curve y = x^2 - 8x + 16

y = x^2 - 8x + 16

y = (x-4)^2 is a perfect square which is always greater than or equal to zero. Y = 0 when x= 4.

y > =  0 for all x.

Therefore y  is minimum when x= 4.

As x increses positively y also increases positively.

Lt x-->infinity  y = Lt x-> infinity (x-4)^2  is also infinity.

Lt x-> -infinity y = Lt x-> -infinity (x-4)^2 = -infinity.

So  y is unbounded  and has no maximum.

giorgiana1976 | Student

The maximum point of these quadratic function is represented by the vertex of the function.

The graph of the quadratic function is a parabola and 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(-b/2a;-delta/4a)=V(4;0)

We notice that the x coordinate is positive and y coordinate is 0, so the vertex of parabola is located on the right side of x axis: V(4;0).

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question