Question

How do you find the axis of symmetry and vertex point of the function: `f(x) = -x^2 + 2x`?

Answer 1

`x=1," vertex "=(1,1)`

Explanation:

`"find the zeros by letting "f(x)=0`

`-x^2+2x=0`

`x(2-x)=0`

`x=0" and "x=2larrcolor(blue)"zeros"`

`"the x-coordinate of the vertex lies on the axis of symmetry"`
`"which is situated at the midpoint of the zeros"`

`"axis of symmetry is "x=(0+2)/2=1`

`"substitute this value into the equation for y-coordinate"`

`y_("vertex")=-1+2=1`

`color(magenta)"vertex "=(1,1)`
graph{-x^2+2x [-10, 10, -5, 5]}

Answer 2

Vertex is at `(1,1)` , axis of symmetry is `x = 1`

Explanation:

`f(x) = - x^2+2 x` or

`f(x) = - (x^2-2 x)` or

`f(x) = - (x^2-2 x+1) +1` or

`f(x) = - (x-1)^2 +1` Comparing with vertex form of

equation `f(x) = a(x-h)^2+k ; (h,k)` being vertex we find

here `h=1 , k=1 :.` Vertex is at `(1,1)`

Axis of symmetry is `x= h or x = 1 ;`

graph{-x^2+2 x [-10, 10, -5, 5]} [Ans]