Question

What is the vertex form of `y= 8x^2+3x-2`?

Answer 1

Vertex `(-3/16,-73/32)`

Explanation:

We will need to complete the square to solve this equation.

First move the constant to the other side of the equation by adding `2` to both sides.

`8x^2+3x=2`

Factor out the coefficient, `8`, from the x^2 term.

`8(x^2+3/8x)=2`

Take the coefficient of the `x` term and divide it by 2 and then square it.

`((3/8)/2)^2=(3/8*1/2)^2=(3/16)^2=9/256`

Add this value to the left hand side

`8(x^2+3/8x+9/256)=2`

Add `8(9/256)` to the right hand side because of the factoring we did earlier.

`8(x^2+3/8x+9/256)=2+8(9/256)`

You now have a perfect square trinomial

`8(x+3/16)^2=2+8(9/256)`

Simplify

`8(x+3/16)^2=2+cancel8(9/(cancel256 32))`

Convert `2` to an improper fraction

`8(x+3/16)^2=64/32+(9/(32))`

Simplify

`8(x+3/16)^2=73/32`

`y=8(x+3/16)^2-73/32`

Vertex form

`y=(x-h)^2+k` where `(h,k)` is the vertex

Vertex `(-3/16,-73/32)`

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