Question 1096091: Y=x^2+8x-2 in vertex form
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! here's a reference.
https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html
you are basically completing the squares as follows:
start with y = x^2 + 8x - 2
add 2 to both sides of the equation to get y + 2 = x^2 + 8x
complete the square on the expression on the right side of the equation to get:
y + 2 = (x+4)^2 - 16
subtract 2 from both sides of the equation to get y = (x+4)^2 - 18
that's your vertex form.
there's another way.
standard form of the equation is y = ax^2 + bx + c
a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.
your equation is y = x^2 + 8x - 2
therefore you get a = 1 and b = 8 and c = -2
the standard form of the vertex form of the equation is y = a * (x-h)^2 + k
h is equal to -b/2a = -8/2 = -4
(x-h) is therefore equal to (x - (-4)) = (x+4)
k is equal to f(h) = (-4)^2 + 8 * (-4) - 2 which is equal to 16 - 32 - 2 which is equal to -16 - 2 which is equal to -18.
a is equal to 1, therefore y = a * (x-h)^2 + k becomes y = 1 * (x+4)^2 - 18 which becomes y = (x+4)^2 - 18.
that's your solution.
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