Question

How do you write the standard form of the equation of the parabola that has (-1/4, 3/2) vertex and (-2,0) point?

Answer 1

`y=-24/49(x+1/4)^2+3/2` or `y=(-24/49)x^2-(12/49)x+72/49`

Explanation:

Use the equation `y=a(x-h)^2+k` where `(h,k)` is the vertex and a is a constant.

The given vertex is `(-1/4,3/2)`.
Plugging this into the equation above gives

`y=a(x--1/4)^2+3/2`

`y=a(x+1/4)^2 +3/2` (EQUATION A)

To find the constant `a`, substitute the given point `(-2,0)` in for `x` and `y`.

`0=a(-2+1/4)^2+3/2`

`0=a(-7/4)^2 +3/2`
`0=a(49/16)+3/2`
`0-3/2=a(49/16)`

`16/49*-3/2=a(49/16)*16/49`
`-24/49=a`

Substitute `a` into equation A above.
`y=a(x+1/4)^2+3/2`
`y=-24/49(x+1/4)^2+3/2`

Some books and teachers call this the standard form; but most call it the vertex form. To find the "usual" standard form, square the binomial and simplify.
`y=-24/49(x+1/4)(x+1/4)+3/2`
`y=-24/49(x^2+1/2x+1/16)+3/2`
`y=(-24/49)x^2 -(12/49)x-3/98+3/2`
`y=(-24/49)x^2-(12/49)x+72/49`