Question

How do you find the VERTEX of a parabola `y= x^2 + 6x + 5`?

Answer 1

Complete the square to get the equation into vertex form:

`y = (x-(-3))^2+(-4)`

Then the vertex can be read as `(-3,-4)`

Explanation:

Given any quadratic: `y = ax^2+bx+c`, we can complete the square to get:

`ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))`

In our example, `a = 1`, `b = 6` and `c=5`, so we find:

`x^2+6x+5 = (x+3)^2 + (5-3^2) = (x+3)^2+(5-9)`

`= (x+3)^2-4`

So `y = (x+3)^2 - 4`

Strictly speaking, vertex form is `y = a(x-h)^2+k`, from which you can read off the vertex `(h, k)`. So let's replace the `(x+3)` with `(x - (-3))` and the `-4` with `+(-4)` to get:

`y = (x-(-3))^2+(-4)`