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]}