Question

How do you graph the inequality `y<=-x^2-3x+10`?

Answer 1

Shade under the parabola.

Explanation:

First, graph the parabola `y=-x^2-3x+10`. If you complete the square, you can rewrite this in vertex form giving you
`y=-(x^2+3x+9/4-9/4-10)`
`y=-((x+3/2)^2-49/4)`
`y=-(x+3/2)^2+49/4`

So the parabola opens downward and has a vertex at
`(-3/2,49/4)=(-1/5, 12.25)`

The `y` intercept occurs when `x=0`, giving `(0,10)`

The `x` intercept occurs when `y=0`, giving `(2,0)` and `(-5,0)`

Finally, if your `y` values are less than the parabola, simply shade underneath the parabola. The graph would look like this: