Question

What is the the vertex of `y = (x -4)^2+9x-20`?

Answer 1

`y = (x +1/2)^2 -17/4`

Vertex `(-1/2 , -17/4)`
Axis of symmetry: `x= -1/2`

Explanation:

Remember quadratic form is `y= ax^2 + bx+c " " " " " " " (1)`

The vertex form of the quadratic equation is `y= a(x-h)^2 + k " " " " " " " " " " " "(2)`

Notice the equation is `y= (x-4)^2 + 9x -20` is not in the vertex form.

We begin by expand the equation like so

`y= (x-4)^2 + 9x -20`

`y = (x-4)(x-4) + 9x - 20`
`y= (x^2 -4x-4x+16)+9x-20`
`y= x^2 -8x + 16 + 9x -20`
`y= x^2 +x -4 " " " " " " " " " " " " " " (3)`

After we simplify function, we have `y = x^2 +x-4` , we can write it in the vertex form by the process of completing the square.

`y = (x^2 + x +color(red)square) -4 -color(blue)(square) " " " " (4)`

Note: the goal of completing the square if to create a perfect trinomial.
The number in the square is `(b/2)^2`

In this case, the middle term is `1` , `(1/2)^2 = 1/4 " " " " " (5)`

`y = (x^2 + x +color(red)(1/4)) -4 -color(blue)(1/4) " " " " (6)`

`y = (x^2 + x +color(red)(1/4)) -16/4 -color(blue)(1/4)`
`y = (x +1/2)^2 -17/4 " " " " " "(7)`

There is alternative method to fid the vertex using
`x_(vertex) = -b/(2a)`

`y_(vertex) = f(-b/(2a))`

I hope this help.