F(x)=x^2-8x+16 x and y intercepts and vertex

So in graphing form F(x)=y so
y=x^2-8x+16
so x intercept is when the line passes through the x axis or when y=0
y intercept is when the line passes throught the y axis or when x=0

x intercept
0=x^2-8x+16
factor
what 2 number multiply to 16 and add to get -8
the numbers are -4 an d -4
0=(x-4)(x-4)
set them to zero
x-4=0
add 4
x=4

x intercept=4 or (4,0)

y intercept
y=x^2-8x+16
x=0
y=0^2-8(0)+16
y=0+16
y=16
y intercept=16 or (0,16)

x intercept is 4 (4,0)
y intercetp is 16 (0,16)

Absolute value graph shifted point 5 up