Also, I thought it said that y=mx+b is a solution for dy/dx, not a tangent line.
@frigidcold Thanks! 
but for the one u said u put | and || …i thought y was changing for the other two graphs. y was not changing for the first one (straight line). there was one that was up and down and the other one was an arc…but dy was
for the average velocity problem on the fr…did they give us velocity or s(t). I think i might have done v(10)-v(0)/10 ;( but i might not have cuz i forgot wat i did but i remember i used the function given…so what function was given? did any take the integral?
@chihuahua1 I believe the question was asking for functions where dy/dx is dependent on y, not those where y wasn’t changing.
And what you did was correct.
On question 6b, did the question ask for the first four non-zero terms of f’ and the general term, or just the first four terms?
@laGrangeError I mean, m is a constant only after you put the value in for the given point. The slope of the differential changes, is what I mean.
@laGrangeError Yes, I think you’re right. I put m=b=1 because I set m=2y-x, and then I substituted the point that we were given in the previous section, (2, 3). However, in hindsight, I don’t think we were supposed to use the point that was given to us in the previous section as it could be done without one (as you showed).
For the dy/dx line thing the lagrange guy is right but you can prove it another way too. If the solution is a line then d2y/dx2 is zero at all points of the solution. Set 2- (2x-y) =0 and you legit get the line y= 2x-2
Graphical analysis saved my life. I probably got 9s on both of the calc parts because of that. The rest were still easy, that series question was a joke. The test was a joke, but I guessed on the optimization mc and I feel bad for the AB people now.
@laGrangeError Now that I think of it, I’m pretty sure part d was redirected to part c. Otherwise, you can get any answers for part d. As abeckuofm223 was saying, (0,0) would change things up. But try another coordinate: it would change things up. That’s because dy/dx is a differential equation rather than a constant term itself. The slope is not a constant but under your assumption, it is.
Correct me if I’m wrong.
@sjwon3789 I was agreeing with Lagrange. Basically the second derivative says that the set of points that satisfy f’'x =0 are related by y=2x-2 which means that y=2x-2 is actually a solution and so m=2 and b=-2
For the Partial fraction decomp one, i don’t remember it too well, but it was (1/6) ln abs (#/x) right? I’m just debating as to whether or not x was in numerator or denominator
Think it was (1/6)ln[(x-6)/x] + c, and this seems right if original question was integral 1/(x^2-6x).
Can anyone else reaffirm this?
ya i think so that was it but i think i forgot to put +c lolol my life
for the tsylor series part b wat did u guys get
i remember my last fourth term was like 9/27 x something
It was just a geometric, multiple by (-3x) each time so: 1, -3x, 9x^2, -27x^3
o i got that wrong then 
wat about number two ik i forgot the speed formula lol so i got that wrong wat were the other ones?
also do any of u guys remember the non calc mc last question that asked about which of the series comverges…wat did u guys put
| and ||?
speed formula, distance was integral of speed formula, slope of tangent was (dy/dt)/(dx/dt) which ended up being just v(y)/v(x), and x position was just integrate x velocity = -.44 and add that to whatever the original x coordinate was
@abeckuofm223 How do you know f’’(x)=0 though? That was the flaw that I pointed out – I’m pretty sure you can’t just say the derivative of m is just 0.
Sjwon i think he’s right, we know f’’(x) = 0 because it said function is y=mx+b which is linear, and for linear function, the 2nd derivative is 0
yea idk if i got that lol thanks! -.44 looks familiar but idk. wat about dp/dt mc one about population …