Question

What is the vertex form of a parabola given vertex (41,71) & zeros (0,0) (82,0)?

Answer 1

The vertex form would be `-71/1681(x-41)^2+71`

Explanation:

The equation for vertex form is given by:
`f(x) = a(x-h)^2 + k` , where the vertex is located at point `(h,k)`

So,substituting the vertex `(41,71)` at `(0,0)`,we get,

`f(x) = a(x-h)^2 + k`

`0=a(0-41)^2+71`

`0 = a(-41)^2 + 71`

`0 = 1681a + 71`

`a = -71/1681`

So the vertex form would be
`f(x) = -71/1681(x-41)^2+71`.