How would you draw the graphs of the interactions between these functions?

Question

I was wondering how I would draw the interactions between these functions if I were not given the equations for them (and am not allowed to make an equation). Please ignore my current drawings. Thanks.

Answer 1

Even if we don't have specific equations for either $f(x)$ or $g(x)$, we can still deduce a decent amount about what $f(x)+g(x)$, $f(x)\cdot g(x)$, and $f(x)/g(x)$ will look like.

$f(x)+g(x)$ is probably the easiest. We know that $g(x)$ must have a negative leading coefficient, because it opens downward, and adding $f(x)$ to it wouldn't change the leading coefficient, because $f(x)$ doesn't have a quadratic term. Therefore, $f(x)+g(x)$ must be a parabola that opens downward. We can also tell that the $y$-intercept of $f(x)+g(x)$ must be less than $0$, since the $y$-intercept of $g(x)$ is more negative than the $y$-intercept of $f(x)$ is positive. Furthermore, we can see that $f(x)+g(x)$ must be greater than $0$ wherever $|f(x)|>|g(x)|$, less than $0$ wherever $|f(x)|<|g(x)|$, and equal to $0$ wherever $|f(x)|=|g(x)|$.

Now we can move on to $f(x)\cdot g(x)$. If the leading coefficient of $f(x)$ is some real number $a>0$ and the leading coefficient of $g(x)$ is some real number $b<0$, then the leading coefficient of $f(x)\cdot g(x)$ must be $ab$ and must be less than $0$, so we know that $f(x)\cdot g(x)$ must be some cubic polynomial that goes to $\infty$ as $x$ goes to $-\infty$ and that goes to $-\infty$ as $x$ goes to $\infty$. We can also tell that the $y$-intercept of $f(x)\cdot g(x)$ must be negative, because the $y$-intercepts of $f(x)$ and $g(x)$ have opposite signs.

$f(x)/g(x)$ is potentially the trickiest, but we can still draw some general conclusions about what it will look like. We know that it will have two vertical asymptotes, one at each value of $x$ such that $g(x)=0$. We also know it will have a horizontal asymptote at $y=0$, because for any large values of $x$, $\frac{1}{g(x)}$ will be extremely close to $0$. In fact, at the one value of $x$ where $f(x)=0$, $f(x)/g(x)$ will be $0$ as well (if both $f(x)$ and $g(x)$ were $0$ at the same place, this wouldn't be the case, but fortunately, we don't have to worry about that). The last piece of information we can determine about $f(x)/g(x)$ is where it is positive and where it is negative: wherever $f(x)$ and $g(x)$ have the same sign, $f(x)/g(x)$ will be positive, and wherever $f(x)$ and $g(x)$ have opposite signs, $f(x)/g(x)$ will be negative. In between those regions of positivity and negativity, $f(x)/g(x)$ must either cross the $x$-axis or have a vertical asymptote.