Question

How do you find the vertex of `y= x^2 + 8x -9`?

Answer 1

`"vertex "=(-4,-25)`

Explanation:

`"given a parabola in "color(blue)"standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

`"then the x-coordinate of the vertex is"`

`•color(white)(x)x_(color(red)"vertex")=-b/(2a)`

`y=x^2+8x-9" is in standard form"`

`"with "a=1,b=8" and "c=-9`

`rArrx_("vertex")=-8/2=-4`

`"substitute this value into the equation for y"`

`y_("vertex")=(-4)^2+8(-4)-9=-25`

`rArrcolor(magenta)"vertex "=(-4,-25)`

Answer 2

The vertex is `(-4,-25)`

Explanation:

You can change the standard form which is given into vertex form which is `y = a(x+p)^2+q` by the process of completing the square. The vertex is `(-p,q)`

`y=x^2 +8x -9`

To complete the square, add and subtract `color(blue)((b/2)^2 =16)`
`y = x^2 +8x " " color(blue)( +16 -16) -9`

`y = (x^2 +8x color(blue)(+16)) +(color(blue)(-16) -9)`

`y = (x+4)^2-25`

The vertex is `(-4,-25)`