How do you graph #y>=-2x^2-x+3#?

1 Answer
Jul 20, 2017

Answer:

I would graph the equation #y = -2x^2-x+3#. This divides the plane into two regions. Test points in each region to see if the region contains solutions to the inequality.

Explanation:

Graph #y = -2x^2-x+3 #.

The graph is a parabola.
We could graph this by finding the vertex and so on, but I find it simpler to just find the #x# intercepts

#-2x^2-x+3 - -(2x^2+x-3) = -(2xcolor(white)"XXX")(xcolor(white)"XXX")#

# = -(2x+3)(x-1)#

So the #x# intercepts are #-3/2# and #1#.

The #y# intercept is #3#, so the graph of the equation is

graph{ -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]}

Testing #(0,0)# we see that #0 >= -2(0)^2-(0)+3# is false so there are no solutions in the region containing the origin.

Testing #(0,5)# (remember the #y# intercept is #3#) or #(1,3)# or #(5,0)# or some other point outside the region containing #(0,0)#, we learn the the region not containing #(0,0)# contains the solutions

OR We can reason the the values of #y# greater than -2x^2-x+3 are above the curve.

graph{y >= -2x^2-x+3 [-8.62, 7.186, -3.96, 3.94]} .