Question

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

Answer 1

The vertex is `(1,-9)`

Explanation:

You have 3 options here:

Option 1

• Multiply out to get the usual form of `y = ax^2 +bx+c`
• Complete the square to get vertex form: `y= a(x+b)^2 +c`

Option 2
You already have the factors.

• Find the roots , the `x`-intercepts. `(y=0)`
• The line of symmetry is halfway between, them this gives `x`
• Use `x` to find `y`. `(x,y)` will be the vertex.

Option 3
- Find the line of symmetry from `x = -b/(2a)`
Then proceed as for option 2.

Let's use option 2 as the more unusual one.

Find the `x`-intercepts of the parabola:

`y= (x-4)(x+2)" "larr` make `y=0`

`0= (x-4)(x+2)" "rarr` gives `x=color(blue)(4) and x= color(blue)(-2)`

Find the midpoint between them: `color(red)(x) = (color(blue)(4+(-2)))/2 = color(red)(1)`

Find the `y`-value using `color(red)(x=1)`

`y= (color(red)(x)-4)(color(red)(x)+2)" "rarr (color(red)(1)-4)(color(red)(1)+2) = -3 xx 3 = -9`

The vertex is at `(x,y) = (1,-9)`