2007-10-11 11:43:55 · 5 answers · asked by LOVELY 1 in Science & Mathematics ➔ Mathematics
y=(x+4)^2-17
-4,-17
2007-10-11 11:49:35 · answer #1 · answered by Kenneth H 3 · 0⤊ 0⤋
You'll need to complete the square to get the equation into the form: y - k = a(x - h)^2 where (h,k) is the vertex.
y = x^2 + 8x -1, take half the 8 and square it, you get 16:
y = x^2 + 8x + 16 -16 -1, now we have a perfect square:
y = (x + 4)^2 - 17 or y + 17 = (x + 4)^2, and we can see that the vertex is (-4,-17).
That's it.
2007-10-11 18:54:17 · answer #2 · answered by Marley K 7 · 1⤊ 0⤋
y = x^2+8x - 1
y+ 1 = x^2 + 8x
y + 1 = (x + 4)^2 - 16
y + 17 = (x+4)^2
vertex (-4, -17)
2007-10-11 18:51:33 · answer #3 · answered by mohanrao d 7 · 0⤊ 0⤋
Differentiate.
y = x^2 + 8x - 1
dy/dx = 2x + 8 = o
x = -4
Vertex is at (-4, -17)
2007-10-11 18:49:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋
y = (x² + 8x + 16) - 16 - 1
y = (x + 4)² - 17
Vertex is (- 4 , -17)
2007-10-15 14:58:25 · answer #5 · answered by Como 7 · 0⤊ 0⤋