Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry of `y=-(x-3)(x-1)`?

Answer 1

Vertex is `(2,1)` and axis of symmetry is `x-2=0`. Graph opens down.

Explanation:

When you write a quadratic equation in form `y=ax^2+bx+c`, it is the coefficient of `a` which tells us whether the graph opens up or down. If `a>0`, it opens up and if `a<0`, it opens down.

When equation is written in vertex form i.e. `y=a(x-h)^2+k`, then `(h,k)` is the vertex and `x-h=0` is the axis of symmetry.

As `y=-(x-3)(x-1)`

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

`=-x^2+4x+3` and as `a=-1<0`, it opens down.

Further `y=-x^2+4x+3`

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

`=-(x-2)^2+1`, which is in vertex form

and vertex is `(2,1)` and axis of symmetry is `x-2=0`

graph{-x^2+4x+3 [-7.875, 12.125, -1.96, 8.04]}