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

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:
$- {x}^{2} + x + 12 = 0$
${x}^{2} - x - 12 = 0$
$\left(x - 4\right) \left(x + 3\right) = 0$
$x = 4 , x = - 3$

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.