Question

How do I convert the equation `f(x)=x^2+2/5x−1` to vertex form?

Answer 1

`color(red)( f(x) = (x+1/5)^2-26/25)`

Explanation:

The vertex form of a quadratic is given by `y = a(x – h)^2 + k`, where (`h, k`) is the vertex.

The "`a`" in the vertex form is the same "`a`" as in `y = ax^2 + bx + c`.

Your equation is

`f(x) = x^2+2/5x-1`

We convert to the "vertex form" by completing the square.

Step 1. Move the constant to the other side.

`f(x)+1 = x^2+2/5x`

Step 2. Square the coefficient of `x` and divide by 4.

`(2/5)^2/4 = (4/25)/4 = 1/25`

Step 3. Add this value to each side

`f(x)+1+1/25 = x^2+2/5x+1/25`

Step 4. Combine terms.

`f(x)+26/25 = x^2+2/5x+1/25`

Step 5. Express the right hand side as a square.

`f(x)+26/25 = (x+1/5)^2`

Step 5. Isolate `f(x)`.

`f(x) = (x+1/5)^2-26/25`

The equation is now in vertex form.

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

`h = -1/5` and `k = -26/25`, so the vertex is at (`-1/5,-26/25`)