# vertex form parabola?

find the vertex form of the parabola

question is y=2x^2+8x-5

choose from these answers (y=2(x+2)^2-9, y=2(x+2)^2-1, y=2(x+2)^2+3, or Y=2(x+2)^2-13)

### 6 Answers

- 1 decade ago
When your function is written in standard form

y = ax^2 + bx + c rather than the vertex form y = a(x - h)^2 + k,

you cannot simply look at the function and find the vertex.

You can find the vertex of your function in two ways:

1-Using the formula h = -b/2a

OR

2-By completing the square

Using the formula is much easier.

You function is y = 2x^2 + 8x - 5.

In your function, a = 2 and b = 8.

Substitute into the formula to find the x-coordinate of the vertex.

h = -(8)/2(2)

h = -8/4

h = -2....This number represents the x-coordinate of your vertex.

Substitute the value of -2 into the original function to get the y-value of the vertex.

y = 2(-2)^2 + 8(-2) - 5

y = 2(4) - 16 - 5

y = 8 - 16 - 5

y = -16 + 3

y = -13

I just found the x and y coordinates, which represent the h and k values in the vertex point (h, k).

Your vertex is (-2, -13).

You got it?

- How do you think about the answers? You can sign in to vote the answer.
- notthejakeLv 71 decade ago
y = 2x^2 + 8x - 5

y = 2(x^2 + 4x + ____) - 5 - 2(___)

y = 2(x^2 + 4x + 4) - 5 - 2(4)

y = 2(x + 2)^2 - 13

the last option...

vertex is (-2, -13), shape factor = 2

- MariaLv 44 years ago
in standard form, the vertex has x value = -b/2a. From that you can find the y value. You now have (h,k) in the vertex form.