What would the equation of this graph be?

Question

Is this a sine graph? How would I model this graph?

Answer 1

It's called interpolating polynomial. There are several methods to find that polynomial based on desired order, number of points to be used, which properties should be conserved and etc. Most straightforward way is to use Lagrange polynomial. For every $n$ distinct points it's possible to construct polynomial of order $n-1$ that passes through those $n$ points. $$ L(x) = \sum_{k=1}^n \ell_k(x) y_k $$ where $$ \ell_k(x) = \prod_{\substack{1 \le m \le n \\m \ne k}}\frac {x-x_m}{x_k-x_m} $$ In your case, you have dataset $$ P = \left \{(0, 0.3), (1, 0.8), (2, 0.8), (3, 0.6), (4, 0.7), (5, 1.7) \right \} $$ You can, and maybe should do it by hands, but I just used Mathematica for that purpose.

pts = {{0, 0.3}, {1, 0.8}, {2, 0.8}, {3, 0.6}, {4, 0.7}, {5, 1.7}};
fLg = InterpolatingPolynomial[pts, x];
pLine = Plot[fLg, {x, 0, 5}, AspectRatio -> Automatic, 
GridLines -> {Automatic, Range[0, 2, 0.1]}, 
PlotRange -> {{0, 5.1}, {0, 2}}, PlotStyle -> {Black, Thick}];
pCircs = Table[Graphics[{Red, Circle[pts[[i]], 0.02]}], {i, 1, 6}];
Show[pLine, pCircs]

which performed `almost' the same graph as you provided

and here's the function itself $$ L(x) = 0.3 + 0.78 x - 0.266667 x^2 - 0.0291667 x^3 + 0.0166667 x^4 - 0.000833333 x^5 $$

Answer 2

This graph cannot represent $\sin$ or $\cos$. Notice how there is one peak and the right-hand side goes higher than the peak.

If you have to guess at what the function might be, look at the ends and the number of troughs/peaks. Ends going in the same direction is how a polynomial of even degree works. Ends going in opposite directions (i.e., like the graph above) is how an odd degree polynomial works.

One peak and one trough probably means that it is a small-degree polynomial, but it is obviously not a straight line, so it must be a minimum of degree 3. More peaks/troughs and/or "flattened" areas of the graph would indicate a higher-degree polynomial.

Since there is an actual peak and trough, it means that $y=x^3$ is not correct, so it must be something like $y=ax^3+bx^2+cx+d$. When $x=0$, $y$ is approximately $0.3$, so I might guess that $d=0.3$. The other values for $a, b, c$ would need further analysis to identify.

Answer 3

Given the data provided on the x'y, can you not plot points at the curve incidents and determine the area of each section. Integrating the equation of the line with respect to y would give the area under the graph, and taking the area under the graph and doing the reverse operation (differentiating with respect to y) will give you the differential. (c = ~0.3 in your case).