How do you graph y=-x^2+x+12?

1 Answer
Sep 11, 2015

The graph looks like this: graph{-x^2+x+12 [-5, 5, -20, 20]}.

Explanation:

Because there is an x^2 term, we know this graph is a parabola.

Because the x^2 term is negative, we know the parabola is in the shape of a downwards U.

First, to find out where the parabola crosses the y-axis, we set x equal to 0 and solve for y:

y = -x^2 + x + 12
y = 12

Next, let's find the two places where the graph intersects the x-axis. To do so, we set the function equal to 0 and factor:
-x^2 + x + 12 = 0
x^2 - x - 12 = 0
(x - 4)(x + 3) = 0
x = 4, x = -3

So the parabola crosses the x-axis and -3 and 4. Now we know know three points for certain:

-- (0, 12) -- where it crosses the y-axis
-- (4, 0) -- one of the x-axis crossing
-- (-3, 0) -- other x-axis crossing

So that should be enough information to draw the graph.