Question

Question #7e641

Answer 1

`f(x)=-2(x-5/2)^2+15/2`

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 constant.

`"to express " f(x)=-2x^2+10x-5" in this form"`

`"Use the method of "color(blue)"completing the square"`

We must ensure that the coefficient of the `x^2` term is 1

`f(x)=-2(x^2-5x)-5larrcolor(red)"coefficient is now 1"`

add`(1/2" coefficient of x-term" )^2 " to " x^2-5x`

`"that is " (-5/2)^2=25/4`

Since we are adding a number we don't have we must also subtract it.

`rArrf(x)=-2(x^2-5xcolor(red)(+25/4)color(red)(-25/4))-5`

`color(white)(rArrf(x))=-2(x-5/2)^2+25/2-5`

`color(white)(rArrf(x))=-2(x-5/2)^2+15/2larrcolor(red)" in vertex form"`