Question

How do you write standard form of quadratic equation of vertex (-1,5) and yint =3?

Answer 1

`y=-2(x+1)^2+5`

Explanation:

The vertex form for a quadratic is
`color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)`
for a parabola with vertex at `(color(red)(a),color(blue)(b))`
`color(green)(m)` indicates if the parabola opens upward (if positive) or downward (if negative) and the "spread" of the parabola.

Since we want a parabola with vertex at `(color(red)(-1),color(blue)(5))`
It must have the form:
`color(white)("XXX")y=color(green)(m)(x+1)^2+5`

So what remains is to determine the value of `color(green)(m)`

W are told that `"yint" = 3` (I assume this means the y-intercept)

Remembering that the y-intercept is the value of `y` when `x=0`
by replacing `y` with `3` and `x` with `0` we have:
`color(white)("XXX")3=color(green)(m)(0+1)^2+5`

`color(white)("XXX")rarr 3=m+5`

`color(white)("XXX")rarr m=-2`

So the complete form for the equation of this parabola is
`color(white)("XXX")y=-2(x+1)^2+5`

Here is a graph of `y=-2(x+1)^2+5` to help verify our result:
graph{(-2)*(x+1)^2+5 [-6.735, 7.31, -0.513, 6.507]}