Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of `y=-(x+1)^2+2`?

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=-(x+1)^2+2" is in this form"`

`"with "h=-1" and "k=2`

`rArrcolor(magenta)"vertex "=(-1,2)`

`"the value of a determines if the graph opens up/down"`

`• " if "a>0" then graph opens up "uuu`

`• " if "a<0" the graph opens down "nnn`

`"here "a=-1rArr" graph opens down"`

`"the axis of symmetry passes through the vertex, is vertical"`
`"with equation"`

`x=-1`
graph{(y+x^2+2x-1)(y-1000x-1000)=0 [-10, 10, -5, 5]}