Question 71295
Some one please help

Determine the vertex of   y = x^2 -8x +22
:
One way to find the vertex is tho put the equation in vertex form by completing the square: {{{highlight(y=(x-h)^2+k)}}}, where (h,k)=(x,y) of the vertex.
y=(x^2-8x+___)-___+22
y=(x^2-8x+(-8/2)^2)-(-8/2)^2+22
y=(x-4)^2-(-4)^2+22
y=(x-4)^2-16+22
y=(x-4)^2+6
The vertex, (h,k)=(4,6)
:
Another way to do this is to use the formula {{{highlight(x=-b/2a)}}} to find the x value of the vertex.
a=1 and b=-8
{{{x=-(-8)/(2(1))}}}
{{{x=8/2}}}
{{{x=4}}}
Now plug that value back into the original equation to find the y-value.
{{{y=(4)^2-8(4)+22}}}
{{{y=16-32+22}}}
{{{y=6}}}
Once again the vertex is (4,6)
Use whichever method you prefer and your teacher allows.
Happy Calculating!!!!