Question

Write f(x)=2x^2+6x-1 in vertex form

Answer 1

y = 2x^2 + 6x -1 ---> y + 1 = 2x^2 + 6x

y + 1 = 2(x^2 + 3x)

Now make a perfect square which has x^2 + 3x as its first two terms. To do this, split the coefficient of the 3x, which is 3, to get 3/2. Square 3/2 to get your constant term, 9/4. Note that when you add 9/4 inside the parentheses, you have to add 18/4 to the other side of the equation, because you have a 2 out in front of the parentheses:

y + 1 + 18/4 = 2(x^2 + 3x + 9/4)

Simplify:

y + 22/4 = 2(x + 3/2)^2

Finally, isolate the y:

y = 2(x + 3/2)^2 - 22/4

It is now in vertex form, y = a(x - h)^2 + k , where h and k are the x and y coordinates of your vertex.

In this case, your vertex is at (h, k) = (-3/2, -22/4)