Question

Question #97c8c

Answer 1

`y=-2(x-1/2)^2+5/2,(1/2,5/2),-58`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`
where (h , k ) are the coordinates of the vertex and a is a multiplier.

`"to obtain this use the method of "color(blue)"completing the square"`

`• " ensure the coefficient of "x^2" term is 1"`

`rArry=-2(x^2-x-1)`

`• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2-x`

`y=-2(x^2+2(-1/2)xcolor(red)(+1/4)color(red)(-1/4)-1)`

`color(white)(y)=-2(x-1/2)^2+5/2larrcolor(red)" in vertex form"`

`rArrcolor(magenta)"vertex"=(1/2,5/2)`

`"when "x=6`

`y=-2(6-1/2)^2+5/2=-121/2+5/2=-58`