Question

What is the vertex form of `y= - x^2 - 10x + 20`?

Answer 1

`y=-(x+5)^2+45`

Explanation:

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

In order to put a parabola into vertex form, use the complete the square method.

`y=-x^2-10x+20`

`y=-(x^2+10x+?)+20`

Add the value that will cause the portion in parentheses to be a perfect square.

`y=-(x^2+10x+25)+20+?`

Since we added `25` inside the parentheses, we must balance the equation.

Notice that the `25` is ACTUALLY `-25` because of the negative sign in front of the parentheses. To balance the `-25`, add `25` to the same side of the equation.

`y=-(x+5)^2+45`

This is the equation in standard form. It also tells you that the vertex of the parabola is `(h,k)`, or `(-5,45)`.

Answer 2

`y=(-xcolor(green)(-5))^2+color(brown)(45)`

Explanation:

By using the vertex form (completing the square) you introduce an error. If this error is '+some value' then you correct by including '- the same value'

Given: `color(blue)(y=-x^2-10x+20)............(1)`

Consider just the right hand side

write as `-1xxcolor(blue)((x^2+10x))+20.........(2)`
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Now consider just the brackets part")`

Write instead : `(x+10/2)^2 ->(x+5)^2`

Multiplying `(x+5)^2` out and you get:

`color(blue)(color(red)((x^2+10x+25) ) <---"Introduced an error of " 25)`

Using this to replace the brackets in expression (2)

`color(blue)(-1xxcolor(red)((x^2+10x+25))+20))`

We have gained the extra value of `color(blue)(-1xx)color(red)( 25)=-25`

so it is `underline(color(red)("NOT CORRECT"))` to write `y= -(x+5)^2+20`

However, it `underline(color(green)("IS CORRECT"))` to write `y= -(x+5)^2color(green)(+25)+20`

Giving the final answer of `color(white)(..)y= -(x+5)^2+45`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`y=(-xcolor(green)(-5))^2+color(brown)(45)`

`color(green)("Notice that "x_("vertex") =-5" as in the brackets")`

`color(brown)("and that "y_("vertex")=45" as the final constant")`