Question 1096091
here's a reference.


<a href = "https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html" target = "_blank">https://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html</a>


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.