Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=-2x^2+8x-3`?

Answer 1

`y`-intercept is `f(0)=-3`, axis of symmetry is `x-2=0` and vertex is `(2,5)`

Explanation:

The vertex form of a quadratic equation is `y=f(x)=a(x-h)^2+k`, where vertex is `(h,k)` and axis of symmetry is `x-h=0` and `y`-intercept is given by `f(0)`.

Writing `f(x)=-2x^2+8x-3` in vertex form, we get

`f(x)=-2x^2+8x-3`

`=-2(x^2-4x+4)+8-3`

`=-2(x-2)^2+5`

Hence `y`-intercept is `f(0)=-3`

axis of symmetry is `x-2=0`

and vertex is `(2,5)`

graph{(y+2x^2-8x+3)(x-2)((x-2)^2+(y-5)^2-0.02)(x^2+(y+3)^2-0.02)=0 [-8.705, 11.295, -3.72, 6.28]}