hrm, you don't need to do that. Your sketching. You know the roots (thats what Ragerunner has said), you know if the leading term is positive/negative and you know if the coefficient of the leading term is odd or even. You don't need to substitute in points. Your sketching, not graphing.
Am i missing something? Only reason I can think of doing something else is if you know there is a double/triple root.
EDIT:
I'll show you what I mean. take:
x^3 - x^2 - 2x > 0 (this is the equation Ragerunner provided
)
according to Ragerunner, we already know the roots, so i'll just say what the roots are.
roots are:
-1, 0 and 2
now, leading term = x^3
therefore graph starts top right (ie, first quadrant at infinity)
as highest power (3) is odd, the graph will end up in the 3rd quadrant.
Now mark the roots.
-1, 0 and 2.
the ONLY way the graph could look is coming down from the top right, passing through 2, turning point below the x axis, going back through 0, turning point above the x axis, and then passing through -1, ending up in the third quadrant at infinity.
As you can see, you dont have to sub in points? I'm sketching the equation, not graphing it, so I don't need to find out where the turning points/inflexion points are.
Hope you see what I mean.
EDIT EDIT:
Oh! If you mean to check whether your graph is right or not, well, yes, you could/should substitute in points, and just double check your graph. But as you can see, its not hard, and there is very little room for error.
