Question

What is the vertex of `y= 4(x+2)^2-2x^2-4x+3`?

Answer 1

`(-3,1)`

Explanation:

Firstly, expand the squared brackets:
`y=4(x^2+4x+4)-2x^2-4x+3`
Then, expand the brackets:
`y=4x^2+16x+16-2x^2-4x+3`
Collect like terms:
`y=2x^2+12x+19`
Use formula for x-turning point: (`-b/{2a}`)
thus, x=`-3`
Plug -3 back into the original formula for y coordinate:
`4(-3+2)^2-2(-3)^2-4(-3)+3=4-18+12+3=1`
therefore the vertex is: `(-3,1)`