Question

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

Answer 1

`"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]}