How do you find the vertex, axis of symmetry and intercepts of f(x)=2x^2+x=3?

1 Answer
Dec 21, 2017

"see explanation"

Explanation:

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

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

"then the x-coordinate of the vertex which is also the"
"axis of symmetry is"

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

f(x)=2x^2+x-3" is in standard form"

"with "a=2, b=1, c=-3

rArrx_(color(red)"vertex")=-1/4

rArr"axis of symmetry is "x=-1/4

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

y_(color(red)"vertex")=2(-1/4)^2+(-1/4)-3=-25/8

rArrcolor(magenta)"vertex "=(-1/4,-25/8)

"for "color(blue)"intercepts"

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

x=0toy=-3larrcolor(red)"y-intercept"

y=0to2x^2+x-3=0

rArr(x-1)(2x+3)=0

rArrx=1" and "x=-3/2larrcolor(red)"x-intercepts"
graph{(y-2x^2-x+3)(y-1000x-250)=0 [-10, 10, -5, 5]}