Find the area of the region determined by

[Cal2isFun]

Find the area of the region determined by the intersections of the curves: y = x³, y=3x+2.
Using the rational root theorem, we find the x intercepts: -1 and 2. We solve and we get the area of the region = 6.75 units.

My question is: say we were to use a graphing calculator to find the intersection, we would only be able to find 1 point of intersection: (x,y) = (2,8).
So, how can we address this obstacle when using a graphing calculator? It simply won't show -1 as an x intercept.
Any ideas?

Thank you!

 

[Inertia_Squared]

There may be a function built into the calculator that you're not aware of, can you give us the model of the calculator that you are using? I'd be glad to go through the manual in my free time and let you know what I find for you

 

[Deleted member 4993]

Find the area of the region determined by the intersections of the curves: y = x³, y=3x+2. Using the rational root theorem, we find the x intercepts: -1 and 2. We solve and we get the area of the region = 6.75 units.

My question is: say we were to use a graphing calculator to find the intersection, we would only be able to find 1 point of intersection: (x,y) = (2,8).
So, how can we address this obstacle when using a graphing calculator? It simply won't show -1 as an x intercept.
Any ideas? Thank you!

Read your user manual (of the calculator) and look up how to change the range of the display - say -5 < x < 5.