# How do you graph the quadratic function y=-x^2+2x+4?

Apr 19, 2015

It's a downward-facing parabola, with a vertical axis.

Let's write the equation as $y = P \left(x\right) = a {x}^{2} + b x + c$

First of all, you notice that $a$ is negative, so you know it's downward-facing.

Then you gotta find the zeros, $P \left(X\right) = 0 \implies$

${X}_{1 , 2} = 1 \pm \sqrt{1 + 16} \implies {X}_{1} = \left(1 + \sqrt{17} , 0\right) , {X}_{2} = \left(1 - \sqrt{17} , 0\right)$

(I know, it's horrible, but it's ok, you know $\sqrt{17}$ is little more than 4, so you can have an idea on where the zeros are)

Now, you know the vertex has ${x}_{v} = - \frac{b}{2} a = 1$,so the vertex v is $\left(1 , P \left(1\right)\right) = \left(1 , a + b + c\right) = \left(1 , 5\right)$

Finally, you know it intercepts the y axis in $\left(0 , P \left(0\right)\right) = \left(0 , c\right) = \left(0 , 4\right)$

So the graph is

graph{-x^2+2x+4 [-20, 20, -10, 10]}