Graphing Rational Functions: Identifying the equation

Question

Graph

I could find the Horizontal & Vertical Asymptotes, Domain, and root, but I still had difficulty finding the equation. I tried utilizing Desmos but fell short. Any ideas of what it might be?

Answer 1

There are several methods we can utilize to narrow down the possible equations for this function. This is an excellent resource for understanding more on how you can find the equation of a polynomial from its graph, but I will try to briefly explain each of those methods.

Any root, or $x$-intercept, of the graph will be a zero of the polynomial. That is, $(x-a)$ will be a factor of a polynomial which has $p(a)=0$ and an $x$-intercept at $x=a$. The multiplicity of this root, or the number of factors of $(x-a)$ that exist for the polynomial, depends on the graph's behavior around $x=a$. If the graph crosses the $x$-axis, the multiplicity is an odd number. You can understand this intuitively because an odd multiplicity means that a change in the sign of $(x-a)$ changes the sign of the product of all the $(x-a)$ terms. Similarly, if the graph is only tangent to the $x$-axis at $x=a$, then the root $a$ has an even multiplicity. Since this is the case with this particular graph, we can know that the function contains an even number of $(x-1)$ terms.

Using similar logic, there must exist more $(x+3)$ and $(x-5)$ terms in the denominator of the function than in the numerator. This is because the graph has vertical asymptotes at $x=-3$ and $x=5$, and that means that the function approaches $-\infty$ as $x$ approaches $-3$ and $5$. However, the $(x+3)$ term is actually a $-(x+5)$ or a $(-x-3)$ term. Think of it in this way. If $x<-3$, $\pm (x+3)>0$, assuming the rest of the function is positive. If $x > -3$, $\pm (x+3)<0$. This means that the term is actually $(-x-3)$ and not $(x+3)$.

Now let's look at the horizontal asymptote. If the numerator and denominator are the same degree polynomials, the numerator divided by the denominator approaches a certain number as $x$ approaches $\infty$ or $-\infty$. This number is the horizontal asymptote. This means that the highest degree term in the numerator is the highest degree term in the denominator times $5$.

Finally, let's talk about the horizontal discontinuity. A discontinuity only happens at a point $c$ if the term $(x-c)$ is there in the numerator and the denominator the same number of times, meaning that they cancel out, leaving only an undefined value at $x=c$. Since this point is at $x=3$, the term $(x-3)$ occurs in the numerator and denominator the same number of times.

Now let's construct a function. Using the above constraints, the function $f(x) = \dfrac{5(x-1)(x-1)(x-3)}{(x+3)(x-5)(x-3)}$ seems like it could work. Graphing this in Desmos gives us a graph that pretty much looks the exact same as your initial graph!

However, it's very important to note that this is only one of many possible functions holding these properties. $g(x) = \dfrac{5(x-1)(x-1)(x-3)\mathbf{(x+3)}}{(x+3)(x-5)(x-3)\mathbf{(x+3)}}$ works just as well, for example. The properties mentioned above hold for all possible functions (the number of $(x-a)$ terms, for example), but the function can be anything where these properties hold true.

PS: Your answer seems to be right, except for the fact that it's $(x-1)^2$ in the numerator, and not $(x+1)^2$ :) Again, I stress on the fact that this is only one of the possible functions, and not the only possible function.