Question

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

Answer 1

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]}