Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of `y=2(x-1)^2-6`?

Answer 1

Vertex is `(1, -6)`

Axis of symmetry `x=1`

Explanation:

Given -

`y=2(x-1)^2-6`

It is like -

`y=a(x-h)+k`
Where `(h, k)` is the vertex.
Since it is in terms of `y` it is either facing up or down.
If `a` is positive, the parabola is facing up.
If `a` is negative, the parabola is facing down.
Axis of symmetry is `x=h`

`h=-(-1)=1`
`k=+(-6)=-6`

Vertex is `(1, -6)`

Axis of symmetry `x=1`

Answer 2

Axis of symmetry is `x=1`, vertex is `(1,-6)` and the graph opens up.

Explanation:

The given equation is in vertex form i.e. `y=a(x-h)^2+k`, whose vertex is `(h,k)` and axis of symmetry is `x-h=0`. Further, if `a<0`, the graph opens down and if `a>0`, the graph opens up.

In the equation `y=2(x-1)^2-6`, we have

Axis of symmetry is `x-1=0` or `x=1`, vertex is `(1,-6)` and as `a>0`, the graph opens up.

graph{(x-1)(y-2(x-1)^2+6)((x-1)^2+(y+6)^2-0.04)=0 [-20, 20, -10, 10]}