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

1 Answer
Andy Y.
May 19, 2017

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:
enter image source here

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