Question

How do you write `y =5x^2 - 3x + 2` in vertex form?

Answer 1

`y=5(x-3/10)^2+31/20`

Explanation:

Vertex form: `y=a(x-h)^2+k`

`y=5x^2-3x+2`

`y=5(x^2-3/5x)+2`

`y=5(x^2-2*3/10x+9/100)+2-5*9/100`

`y=5(x-3/10)^2+2-9/20`

`y=5(x-3/10)^2+31/20`

Answer 2

`y= 5(x-3/10)^2 +31/20`

Explanation:

Vertex form is a useful way of writing the equation of a parabola because it tells you the co-ordinates of the turning point, (the vertex).

It is written as `y = p(x+q)^2 +r`
The vertex is at `(-q,r)`

`y = 5x^2-3x+2" "larr` in the form `ax^2+bx+c`

`y = 5[x^2-3/5x" "+2/5]" "larr` you must have `1x^2`

Now you need to add in a term to create the square of the binomial.
The missing term comes from `(b/2)^2` where `b = (-3/5)`

`(-3)/5 div 2 =(-3)/10`

If you add an extra term, you must subtract it as well to keep the equation the same.

`y = 5[x^2-3/5x + color(green)(((-3)/10)^2)+2/5 -color(green)(((-3)/10)^2]`

`y = 5[x^2-3/5x +9/100+2/5 -9/100]`

`y = 5[(color(blue)(x^2-3/5x +9/100))+(color(red)(2/5 -9/100))]`

`y = 5[color(blue)((x-3/10))^2+(color(red)(40/100 -9/100))]`

`y = 5[color(blue)((x-3/10))^2+(color(red)(31/100))]" "larr` distribute the `5`

`y = 5color(blue)((x-3/10))^2+5(color(red)(31/100))`

`y= 5(x-3/10)^2 +31/20`

The vertex is `(3/10,31/20)`