Find the equation of the function graphed below.

Find the equation of the rational function from the image.

I know that there are vertical asymptotes at $$x=-1$$ and $$x=2$$. There are also x intercepts at $$x=1$$ and $$x=3$$. Since the horizontal asymptote is at $$y=1$$, the degrees of the numerator and denominator have to be the same. Thus I got $$f(x)= \frac{(x-1)^3(x-3)}{(x+1)^2(x-2)^2}.$$

However, the y-intercept of graph is at $$(0,1)$$ while the function I created has a y-intercept at $$(0,\frac{3}{4})$$. How do I change my function to make it have a y-intercept of $$(0,1)$$?

Also, I am not sure if the cube is supposed to be attached to the $$(x-1)$$ binomial. I just stuck the cube to any binomial to make the degrees of the numerator and denominator the same. How do I know where the cube is supposed to go?

• It is simpler to handle separately the horizontal asymptote by looking for a form $y=1+$something where "something" is tending to $0$ at infinity . Coping with the vertical asymptotes and the different signs, I find $y=1+ax(x-1)(x-3)/((x+1)(x-2))^2$ for any positive value of $a$. Commented Sep 22, 2022 at 6:57
• @JeanMarie Wouldn't the graph of that equation cross the line $y=1$ at $x=1$ and $x=3$ rather than crossing the $x$ axis at $x=1$ and $x=3$? Commented Sep 22, 2022 at 13:26
• @David K My bad. You are perfectly right. Commented Sep 23, 2022 at 13:51

The only mistake you made was assuming that you had to cube one of the factors in the numerator to make the degrees of the numerator and the denominator the same. In fact, if you look at how the graph crosses the $$x$$-axis at $$1$$ and $$3$$, you'll see neither of the two factors in the numerator should be cubed, otherwise you'd see an inflection point (i.e. the graph would "flatten") at one of those two x-intercepts.
Instead of that, you need to add another factor in the numerator without adding to the number of x-intercepts; the simplest one would have the form $$\,(x^2+a)\,$$ for some positive $$\,a\,$$. Now you just need to figure out the right value for $$\,a\,$$ to make your $$\,y$$-intercept be $$1$$.