Question

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

Answer 1

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.