Question

How do you find the vertex of `y=2x^2+10x+8`?

Answer 1

`(-5/2,-9/2)`

Explanation:

`"Given the equation of 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=2x^2+10x+8" is in standard form"`

`"with "a=2,b=10" and "c=8`

`rArrx_("vertex")=-10/4=-5/2`

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

`y_("vertex")=2(-5/2)^2+10(-5/2)+8=-9/2`

`rArrcolor(magenta)"vertex "=(-5/2,-9/2)`
graph{2x^2+10x+8 [-10, 10, -5, 5]}