2007-06-14 02:51:53 · 5 answers · asked by Shorty 2 in Science & Mathematics ➔ Mathematics
y = x^2 - 8x + 22
. . . -8/2 = -4
. . . -4^2 = 16
y = (x^2 - 8x + 16) + 22 - 16
y = (x - 4)^2 + 6
The vertex is at (4, 6)
2007-06-14 02:56:04 · answer #1 · answered by TychaBrahe 7 · 0⤊ 0⤋
Take the derivative of the equation:
y' = 2x - 8
Set the derivative to 0:
0 = 2x - 8
Solve for x:
x = 4
Plug back in to the original:
y = 4^2 - 8(4) + 22
y = 6
So, the vertex is at (4,6). Just remember that is you take a derivative and set it to 0, you are finding a minimum or maximum value of the function. This is where the vertex is located.
2007-06-14 03:00:43 · answer #2 · answered by yeeeehaw 5 · 0⤊ 0⤋
Differentiating and setting equal to zero I get 2x = 8, or x = 4
putting that value of x into the original eq, y = 16 - 32 +22 = 6
2007-06-14 02:58:09 · answer #3 · answered by Steve 7 · 0⤊ 0⤋
dy/dx = 2x - 8 = 0 => x = 4
4² -8*4 + 22 = 6
vertex at (4, 6)
Doug
2007-06-14 02:59:01 · answer #4 · answered by doug_donaghue 7 · 0⤊ 0⤋
y=x^2-8x+16+6=(x-4)^2+6 so the vertix is(4,6)
2007-06-14 02:59:01 · answer #5 · answered by santmann2002 7 · 0⤊ 0⤋