I am very much confused on finding the slope the curve at a specific point on a graph. I know the process, draw a tangent at the point and then estimate $y$ units per $x$ units. I am doing that but it doesn't match with the answer sheet. Different people will have different tangent line with same point, isn't it? My graph is attached. What is the slope at the point P1?

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    $\begingroup$ You should include your work in order for us to help. In particular, what tangent line did you draw? What slope did you get? Also, what does the answer sheet say? $\endgroup$ – Dave Jun 23 '18 at 4:15
  • $\begingroup$ Not just work. They actual question and answer. What point are you trying to do and why don't you get it right. How the heck are we supposed to know why you got it wrong? $\endgroup$ – fleablood Jun 23 '18 at 5:37
  • $\begingroup$ Why do you refuse to tell us the answer you got and the answer in the answer sheet? $\endgroup$ – Joel Reyes Noche Aug 3 '18 at 5:17

Approximations are just approximations. I would say the slope at $P_1$is approximately $-3$

To check my approximation I found the equation for the cubic curve passing through $$(-1,-2), (0,0),(1,2),(-2,2)$$

I found $y= -x^3+3x$ whose derivative at $P_1$ is $-\frac {7}{3}$

  • $\begingroup$ Thanks RiaziKeramani Answer sheet was not giving good approximation & it was confusing me $\endgroup$ – user3352074 Jun 29 '18 at 13:50
  • $\begingroup$ @user3352074 Thanks for your attention. These approximation problems are often not well posed. You need to find some points which are on the curve and find the coordinates from the graph. $\endgroup$ – Mohammad Riazi-Kermani Jun 29 '18 at 14:35

Try the calculus method. Take a point $K_1$ near $P_1$ then draw a secant $P_1K_1$. Then take a point $P_2$ between $P_1$ and $K_1$, and draw $P_1K_2$. Continue this process until $K_n$ is very close to $P_1$. It'll be a good approximation of the tangent.


  • $\begingroup$ Thanks I know this process. $\endgroup$ – user3352074 Jun 29 '18 at 13:51

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