So, this assignment was daunting at first... We had this complex moving graph and we had to recreate it. Once our group thought about the little tasks we had to accomplish and got to work at taking them out, things began to come easier.
The first major realization was that there was always going to be one fixed point on our new line and that point was going to be (2,2). So that helped us to fill in the blanks of what to put into our point slope formula.
Next we needed to figure out the slope. Well, the slope formula is (y2-y1)/(x2-x1) so we put in our known points in. (y2-2)/(x2-2) and then we figured that the slope is h,f(h) so we put in ((.5h^2-2)/(h-2))(x-h)(.5h^2) then it would work. It did work.
here is a little gif/video of the equation we found
https://www.youtube.com/watch?v=FZHM6tyBVY8&index=2&list=UUWw0Umy_GSQUYf41aYt8zsA
For the second equation all that changes is the (2,2) stationary point, so we made a slider for that (c) and there you go.
https://www.youtube.com/watch?v=mcANOuK9HHI&index=1&list=UUWw0Umy_GSQUYf41aYt8zsA
So, to find the tangent of a line you find the limit of the slope when the equation approaches 0 and then plug that in for the slope and you are set. Here is a bonus graph we made where it is tangent everywhere. Enjoy.
https://www.youtube.com/watch?v=MotPL189qYc&index=3&list=UUWw0Umy_GSQUYf41aYt8zsA
The first major realization was that there was always going to be one fixed point on our new line and that point was going to be (2,2). So that helped us to fill in the blanks of what to put into our point slope formula.
Next we needed to figure out the slope. Well, the slope formula is (y2-y1)/(x2-x1) so we put in our known points in. (y2-2)/(x2-2) and then we figured that the slope is h,f(h) so we put in ((.5h^2-2)/(h-2))(x-h)(.5h^2) then it would work. It did work.
here is a little gif/video of the equation we found
https://www.youtube.com/watch?v=FZHM6tyBVY8&index=2&list=UUWw0Umy_GSQUYf41aYt8zsA
For the second equation all that changes is the (2,2) stationary point, so we made a slider for that (c) and there you go.
https://www.youtube.com/watch?v=mcANOuK9HHI&index=1&list=UUWw0Umy_GSQUYf41aYt8zsA
So, to find the tangent of a line you find the limit of the slope when the equation approaches 0 and then plug that in for the slope and you are set. Here is a bonus graph we made where it is tangent everywhere. Enjoy.
https://www.youtube.com/watch?v=MotPL189qYc&index=3&list=UUWw0Umy_GSQUYf41aYt8zsA