Question

How do write standard form of equation of the parabola given that vertex is (5/2, - 3/4) and graph passes through point (-2 ,4)?

Answer 1

Express the parabola in vertex form and work from there to obtain `y=19/81x^2-95/81x+58/81`.

Explanation:

We might just want to jump into the problem and try finding the standard form `y=ax^2+bx+c` of this parabola, but we don't have enough information to do this yet. Instead, we have to use the vertex form of a parabola:
`y=a(x-h)^2+k`
Where `(h,k)` is the vertex.

The vertex is given in the problem, so:
`y=a(x-5/2)^2-3/4`

We have to find the value of the constant `a` (which determines how thin or thick the parabola is), and we can do this using the point `(-2,4)`:
`y=a(x-5/2)^2-3/4`
`->4=a(-2-5/2)^2-3/4`
`->4+3/4=a(-9/2)^2`
`->19/4=81/4a`
`->a=19/81`

Our equation in vertex form is therefore:
`y=19/81(x-5/2)^2-3/4`

We can expand the right hand side to find the standard form:
`y=19/81(x-5/2)^2-3/4`
`->y=19/81(x^2-5x+25/4)-3/4`
`->y=19/81(x^2-5x+25/4)-3/4`
`->y=19/81x^2-95/81x+475/324-3/4`
`->y=19/81x^2-95/81x+58/81`