Question 969032
all the graph doesn't have to be above the x-axis.


they are looking for the parts of the graph that are above the x-axis.


the x-axis is when y = 0.


you solve for y = ax^2 + bx + c > 0 by setting y = ax^2 + bx + c = 0, and then finding the area of the graph that are above the x-axis.


here's an equation.


y = x^2 + x - 6.


if they ask you to find wyhen x^2 + x - 6 is greater than 0, you would first find when x^2 + x - 6 is equal to 0 and then look for the areas of the graph that are above the x-axis.


in this problem, you would do the following.


you would find the roots.


they would be at x = 2 and x = -3.


you would then determine the regions of the graph.


they would be x < -3, -3 < x < 2, x > 2.


you would then test each region to see if the value of y is positive or negative.


your eqution is: y = x^2 + x - 6.


when x = -5, y is equal to 14.


when x is equal to 0, y is equal to -6


when x is 5, y is equal to 24.


the graph is positive when x < - 3.
the graph is negative when -3 < x < 2.
the graph is positive when x > 2.


that would be your solution.


here's what that graph looks like:


{{{graph(400,400,-10,10,-50,50,x^2 + x - 6)}}}


you can see from the graph that x^2 + x - 6 > 0 in the intervals we just calculated.


here's a reference on polynomial inequalities you might find useful.


<a href = "http://home.windstream.net/okrebs/page32.html" target = "_blank">http://home.windstream.net/okrebs/page32.html</a>


here's another one.


<a href = "http://www.regentsprep.org/regents/math/algtrig/ate6/quadinequal.htm" target = "_blank">http://www.regentsprep.org/regents/math/algtrig/ate6/quadinequal.htm</a>