This is also the graph of a parabola only it is in the more general form.

\[y = a{x^2} + bx + c\]

In this form, the \(x\)-coordinate of the vertex is \(x = - \frac{b}{{2a}}\) and we get the \(y\)-coordinate by plugging this value back into the equation. So, for our parabola the coordinates of the vertex will be.

\[\begin{align*} x & = - \frac{2}{{2\left( { - 1} \right)}} = 1\\ y & = - {\left( 1 \right)^2} + 2\left( 1 \right) + 3 = 4\end{align*}\]

So, the vertex for this parabola is (1,4).

We can also determine which direction the parabola opens from the sign of \(a\). If \(a\) is positive the parabola opens up and if \(a\) is negative the parabola opens down. In our case the parabola opens down.

This also means that we’ll have \(x\)-intercepts on this graph. So, we’ll solve the following.

\[\begin{align*} - {x^2} + 2x + 3 & = 0\\ {x^2} - 2x - 3 & = 0\\ \left( {x - 3} \right)\left( {x + 1} \right) & = 0\end{align*}\]

So, we will have \(x\)-intercepts at \(x = - 1\) and \(x = 3\). Notice that to make my life easier in the solution process I multiplied everything by -1 to get the coefficient of the \({x^2}\) positive. This made the factoring easier.

Here’s a sketch of this parabola.

We could also use completing the square to convert this into the form used in the previous problem. To do this I’ll first factor out a -1 since it’s always easier to complete the square if the coefficient of the \({x^2}\) is a positive 1.

\[\begin{align*}y & = - {x^2} + 2x + 3\\ & = - \left( {{x^2} - 2x - 3} \right)\end{align*}\]

Now take half the coefficient of the \(x\) and square that. Then add and subtract this to the quantity inside the parenthesis. This will make the first three terms a perfect square which can be factored.

\[\begin{align*}y & = - \left( {{x^2} - 2x - 3} \right)\\ & = - \left( {{x^2} - 2x + 1 - 1 - 3} \right)\\ & = - \left( {{x^2} - 2x + 1 - 4} \right)\\ & = - \left( {{{\left( {x - 1} \right)}^2} - 4} \right)\end{align*}\]

The final step is to multiply the minus sign back through.

\[y = - {\left( {x - 1} \right)^2} + 4\]