Question

What is the vertex form of `y=(5x-5)(x+20)`?

Answer 1

vertex form: `y=5(x+19/2)^2-2205/4`

Explanation:

1. Expand.
Rewrite the equation in standard form.

`y=(5x-5)(x+20)`

`y=5x^2+100x-5x-100`

`y=5x^2+95x-100`

2. Factor 5 from the first two terms.

`y=5(x^2+19x)-100`

3. Turn the bracketed terms into a perfect square trinomial.
When a perfect square trinomial is in the form `ax^2+bx+c`, the `c` value is `(b/2)^2`. So you have to divide `19` by `2` and square the value.

`y=5(x^2+19x+(19/2)^2)-100`

`y=5(x^2+19x+361/4)-100`

4. Subtract 361/4 from the bracketed terms.
You can't just add `361/4` to the equation, so you have to subtract it from the `361/4` you just added.

`y=5(x^2+19x+361/4` `color(red)(-361/4))-100`

5. Multiply -361/4 by 5.
You then need to remove the `-361/4` from the brackets, so you multiply it by your `a` value, `color(blue)5`.

`y=color(blue)5(x^2+19x+361/4)-100[color(red)((-361/4))*color(blue)((5))]`

6. Simplify.

`y=5(x^2+19x+361/4)-100-1805/4`

`y=5(x^2+19x+361/4)-2205/4`

7. Factor the perfect square trinomial.
The last step is to factor the perfect square trinomial. This will tell you the coordinates of the vertex.

`color(green)(y=5(x+19/2)^2-2205/4)`