Question

How do you write the vertex form equation of the parabola `y=-x^2+2x+3`?

Answer 1

Complete the square.

Explanation:

For `y=-x^2 + 2x +3`, take out the factor of `-1` to make the `x^2` coefficient `+1`.

`:. y=-(x^2-2x-3)`

Now, complete the square. Divide the `x` coefficient by `2` and square it, adding it and subtracting it.

`:. y=-(x^2-2x+1-1-3)`
`:. y=-((x-1)^2-4)`
`:. y=-(x-1)^2 + 4`

This is now in turning-point form; simply read off the coordinates of your vertex, which is at `(1,-4)`. Remember that this is a local maximum. Of course, the original equation factorises quite easily, so for sketching purposes, make `y=-(x-3)(x+1)` to get your `x`-intercepts; the `y`-intercept is, of course, just the constant term in the original equation.