Interpreting equations graphically (example 2) | Mathematics III | High School Math | Khan Academy

Let F of T be ( e^{2T} - 2T^2 ) and H of T be ( 4 - 5T^2 ). The graphs of Y = F(T) and Y = H(T) are shown below. So, Y = F(T) is here in green, so this is really ( Y = e^{2T} - 2T^2 ). We see F(T) right over there, and Y = H(T) is shown in yellow.

Alright, now below that they say which of the following appear to be solutions of ( e^{2T} - 2T^2 = 4 - 5T^2 )? Select all that apply, and I encourage you to pause the video and try to think about it.

Now, the key here is to realize that ( e^{2T} - 2T^2 ) that was F(T) and ( 4 - 5T^2 ) is H(T). So another way of thinking about it: select all of the T's for which F(T) is equal to H(T). So all of the T's where F(T) is equal to H(T, well that's going to happen at the points of intersection.

For example, at T1, we see at this point right here T1, ( Y1 ). So this tells us ( F(T1) = H(T1) ), which is equal to ( Y1 ). So F(T) is going to be equal to H(T) at T = T1, and we see that there because it's a point of intersection.

Now let's keep on going. Well, they have another point of intersection right over here at T4, T4, ( Y4 ). If you took F(T4), you're going to get ( Y4 ), or if you take H(T4), you're going to get ( Y4 ). So ( F(T4) = H(T4) ).

Thus, ( F(T4) = H(T4) ). If you took ( e^{2 \cdot T4} - 2T4^2 ), that is going to be equal to ( 4 - 5 \cdot T4^2 ). So ( T4 ), since it satisfies both F(T) and H(T), equals each other when T is equal to T4.

These two things are going to equal each other when T is equal to T4, and those are the only ones that are at a point of intersection. I think we are done. Check my answer, and got it right.