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

Jun 4, 2017

Complete the square.

#### Explanation:

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

$\therefore y = - \left({x}^{2} - 2 x - 3\right)$

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

$\therefore y = - \left({x}^{2} - 2 x + 1 - 1 - 3\right)$
$\therefore y = - \left({\left(x - 1\right)}^{2} - 4\right)$
$\therefore y = - {\left(x - 1\right)}^{2} + 4$

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