Where is the vertex of the graph of f(x) = x^2 - 8x + 16 ?
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
delta =64 - 64
delta = 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).