Question

Given the vertex of a parabola is (3,4) and the parabola goes through the (-2,5), find the general form of the parabola?

Answer 1

`y=1/25x^2-6/25x+109/25`

Explanation:

I'm assuming you mean the standard form of this parabola:

Let's start with the vertex form of the parabola:
`y=a(x-h)^2+k`

You already have the vertex, so substitute the `(h,k)`
`y=a(x-3)^2+4`

Now to find `a`, substitute in the point you have been given:
`5=a(-2-3)^2+4`
`5=a(-5)^2+4`
`5=25a+4`
`1=25a`
`1/25=a`

Now substitute the `a` and the `(h,k)` and solve for the standard form:
`y=1/25(x-3)^2+4`
`y=1/25(x^2-6x+9)+4`
`y=1/25x^2-6/25x+9/25+4`
`y=1/25x^2-6/25x+109/25`