Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of `y=-4(x-2)^2+2`?

Answer 1

Given the vertex form `y = a(x-h)^2+k" [1]"`

If `a >0`, then the graph opens up
If `a<0`, then the graph opens down.
The vertex is the point `(h,k)` and the axis of symmetry is `x = h`

Explanation:

For the given equation, `y=-4(x-2)^2+2`, we observe that `a = -4, h = 2 and k =2`

-4 < 0, therefore, the graph opens down.

The vertex is the point `(2,2)`

The axis of symmetry is `x = 2`

Answer 2

`"down " (2,2),x=2`

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 constant.

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

`"if " a<0" then graph opens down " nnn`

`"here " a=-4rArr" graph opens down"`

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

`h=2" and " k=2`

`rArrcolor(magenta)"vertex "=(h,k)=(2,2)`

The axis of symmetry is vertical and passes through the vertex

`rArr" equation of axis of symmetry is " =2`
graph{(y+4x^2-16x+14)(y-1000x+2000)=0 [-10, 10, -5, 5]}