Question

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

Answer 1

`(-3,0),-27,-3`

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+3)^2" is in vertex form"`

`"with "h=-3" and "k=0`

`rArrcolor(magenta)"vertex "=(-3,0)`

`color(blue)"to find intercepts"`

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

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

`x=0toy=-3(9)=-27larrcolor(red)"y-intercept"`

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

`rArr(x+3)^2=0`

`rArrx+3=0rArrx=-3larrcolor(red)"x-intercept"`
graph{-3(x+3)^2 [-10, 10, -5, 5]}