How do you determine a function statement based on a graph?

Answer

Dear Ess,

the figures are ambiguous to define a unique function that looks like this, and moreover the axes are rather crude.

So I’m going to have to make some assumptions and limit myself to some general guidelines.

To begin with, every point in the plane has an x ​​coordinate and a y coordinate. The lines you see in the figure link a y-coordinate to each x-coordinate: if you know the x-coordinate, you can find the (unique) y-coordinate associated with that x-value. We can write that calculation as y=f(x). The function f tells you what calculations to perform on x to get y. If you do that calculation for each value of x, you get the corresponding value of y. That collection of points lies on a curve, and that’s what you see drawn.

Now, in the problem you present here, you get that curve, and you’re asked to formulate a “reasonable” f that gives a similar curve. You will then have to limit yourself to a certain class of functions, and I suspect you only want to consider functions that are polynomials, or rational functions:

• a polynomial has the form y = f(x) = a*x^n + b*x^{n-1} +…. + d, where n is a natural number and a, b, …, d also known numbers.
• a rational function has the form y = f(x) = p(x)/q(x) with p(x) and q(x) polynomials.

Now, what can we know about this kind of function based on above figure. How to do this technically is well explained in every math textbook on the subject and in many places on the internet. I limit myself to the principles: what are we trying to do, in what order, and why. If you know that, you will be able to find the effects.

1) The first thing to look at is the zeros. These are those values ​​of x for which y=0. In the left figure these are 0, 1, 2, and 3. So we need to find a polynomial that becomes zero at these points. A convenient way to do that is to write y = a * x * (x-1) *(x-2)*(x-3). If you expand this product, you get a polynomial of the form given earlier.

You notice that we can multiply that polynomial by any number a, and that the zeros don’t change as a result. You can try to deduce what a is from the height of the maxima. You then calculate the derivative of your polynomial and see where it becomes 0. The x value for which the derivative becomes 0 is a maximum of the original function. You can then evaluate the original function at that point, and choose a such that the value of the function at the maximum equals 1 (which seems reasonable to me based on the figure).

This is enough for the figure on the left. There is no reason to believe that function is not a polynomial.

2) The right figure is a bit more complicated. You see that function is discontinuous and infinite at certain points. We call these vertical asymptotes. If we assume a rational function f(x)=p(x)/q(x) , then you can easily see that such a rational function becomes infinite at a zero of the denominator (which is not a zero of the numerator ). In that case you get “something finite” divided by 0.

I cannot deduce from the figure exactly where those vertical asymptotes lie, but x = 1 and x = -1 seem like a reasonable guess.

So we now know that f(x) = p(x)/(x^2-1), because its denominator becomes 0 for x = 1 and x=-1, and we haven’t said anything about the numerator yet.

Now that you know this, we can get started for the counter. We see that x=0 is a zero, and even a multiple zero, since the function appears to touch the x-axis at 0 (and thus does not change sign there). So the counter already contains a factor of x^2. In addition, we also see that for very large x-values ​​and very small x-value the function starts to resemble a straight line, so of the form

y=a*x+b.

You could derive the values ​​a and b for a good straight line that closely matches the figure. They are known.

That line is called a slanted asymptote, and you need to make sure that your function f(x)=p(x)/(x^2-1) gets closer and closer to that line as x grows. So you demand that the limit for x going to infinity corresponds to that line you already have. You can find out how to do this technically in many places on the internet.