Question

How do you find the vertex and the intercepts for `y = -3(x - 2)^2 + 4`?

Answer 1

`"see explanation"`

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 multiplier.

`y=-3(x-2)^2+4" is in vertex form"`

`"with "(h,k)=(2,4)larrcolor(red)" vertex"`

`"to find the intercepts"`

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

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

`x=0toy=-3(-2)^2+4=-8larrcolor(red)" y-intercept"`

`y=0to-3(x-2)^2+4=0`

`rArr-3(x-2)^2=-4`

`rArr(x-2)^2=4/3`

`color(blue)"take the square root of both sides"`

`rArrx-2=+-sqrt(4/3)larr" note plus or minus"`

`rArrx=2+-2/sqrt3=2+(2sqrt3)/3`

`rArrx~~ 0.85" or "x~~ 3.15larrcolor(red)" x-intercepts"`
graph{-3(x-2)^2+4 [-10, 10, -5, 5]}