The main question is in the title but here is some background. I am writing a quiz for my Algebra 2 class for which they have to graph a parabola (ex. $7x^2+3x-2$) for question 1 and graph a rational function whose numerator and denominator are linear for question two (ex. $\frac{3x-1}{x+5}$). But then I thought it would be fun for an extra credit question to ask them if they could find the points where these functions crossed. Thus they must equate a parabola to a rational function: $$ax^2+bx+c=\frac{dx+e}{fx+g}$$ which is where I got the equation in the title (obviously $a$,$f$ and $g$ are nonzero). From contemplation, the nicely behaved functions of this nature will cross at three distinct points, which I hope for in the construction of the problem. My overall goal was for this to also be an echo of what we learned when solving polynomial equations; specifically the rational root theorem. I’m just trying to make the numbers work out nicely for them. Any advice here?
For what constants $a,b,c,d,e,f,g\in\mathbf{Z}$ is $(ax^2+bx+c)(fx+g)-(dx+e)=0$ solvable via the *Rational Root Thm.*
algebra-precalculuslinear algebrarational numbers
Best Answer
I'd choose the rational function first. Make it anything you want, but keep the numbers reasonable. You'd like for there to be three points on it with integer coordinates, which isn't too hard to accomplish. Suppose you go with, just choosing randomly here: $$r(x)=\frac{3x+4}{x-1}$$ You'll certainly get integer outputs if $x=0$ or $x=2$. You don't want $0$ to be one of your roots, though, or it's not a good rational root theorem problem. Trying a few values, it appears that $x=-6$ and $x=8$ also work.
Now you just need a parabola that has the same values as $r(x)$ at the points we chose. That is, we want $p(x)=ax^2+bx+c$, where
You can use interpolate a parabola that passes through the points $(-6,2)$, $(2,10)$, and $(8,4)$ using the equations:
$\begin{cases}36a-6b+c=2 \\ 4a+2b+c=10 \\ 64a+8b+c=4\end{cases}$
Solving this, you get a quadratic with rational coefficients: $-\frac17x^2+\frac37x+\frac{68}7$
That's kind of gross, so go ahead and write down your equation, "parabola = hyperbola", and then multiply both sides by, say, $-7$:
$$-\frac17x^2+\frac37x+\frac{68}7=\frac{3x+4}{x-1} \\ \implies x^2-3x-68=\frac{21x+28}{1-x}$$
As you can see from the graph (https://www.desmos.com/calculator/4d7q5qsyzj), the curves do cross at the appropriate points.
Your students, after multiplying the denominator across and collecting terms to one side, will obtain the cubic: $$x^3-4x^2-44x+96=0$$
Although $96$ looks a little intimidating, it only has twelve factors. If they try $x=2$ as an early guess, then they'll be down to a very tractable quadratic.
You could probably also be a bit more intentional about choosing where the roots will be. Suppose you want them at $x=-1,3,5$. Just write your rational function with odd coefficients on top, and $x-1$ in the denominator again. That guarantees that you get $\dfrac{\text{even}}{\pm 2}$ when $x$ is $-1$ or $3$, and then just choose $e$ so that that $d(5)+e$ is a multiple of $4$. A simple one that works would be $$\frac{3x+1}{x-1},$$ which goes through the points $(-1, 1), (3, 5), (5,4)$. The parabola passing through those points is $y=-\frac14x^2+\frac32x+\frac{11}4$.
Again, equate the two, and multiply by $-4$ this time, and then see the resulting cubic:
$$-\frac14x^2+\frac32x+\frac{11}4=\frac{3x+1}{x-1} \\ \implies x^2 - 6x - 11 = \frac{12x+4}{1-x} \\ \implies x^3-7x^2 +7x+15=0$$
That's not a bad little rational roots exercise!
I hope this answer has given you an idea of what kind of tinkering and playing around is necessary to make a problem like this "work", with numbers that you feel are appropriate for your students.