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

1 Answer
Lili D.
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.

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