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

1 Answer
Apr 19, 2015

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

Let's write the equation as y=P(x)=ax^2+bx+c

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

Then you gotta find the zeros, P(X)=0 =>

X_{1,2}=1+-sqrt(1+16) => X_1=(1+sqrt(17),0), X_2=(1-sqrt(17),0)

(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= -b/2a=1,so the vertex v is (1,P(1))=(1,a+b+c)=(1,5)

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

So the graph is

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