# Where is the vertex of the graph of f(x) = x^2 - 8x + 16 ?

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f(x) = x^2 -8x+16. To find the vertex of the graph.

Let y = x^2-8x+16

y = x^2-8x+4^2

y = (x-4)^2.

Or (x-4)^2 = 4(1/4)(y-0).

If we compare (x-4)^2 = 4*(1/4)(y-0) to the standard parabola (x-h)^2 = 4a(y-k) whose vertex is (h,k), we get:

the vertex of (x-4)^2 = 4*(1/4)(y-0) is at (4,0).

So the vertex of f(x) = x^2-8x+16 is at (4, 0).

To discover where the vertex of the parable f(x) = y is located, we'll have to establish the quadrant or the side of x axis where the coordinates of the vertex of the graph of y are located.

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) **

**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).**