# sketch the graph of a function with constants

Sketch the graph of $$f(x)=1+\frac ax+\frac {a} {x^2}$$, $a \gt0$

how is it even possible to draw a graph with constant? I mean how can I sketch it? it with different values of a my graph and its asymptotes will change, at least the horizontal one.
ok, here I can say for sure that vertical asymptote is $x=0$. But how to proceed with critical and inflection points? how can I tag them on graph?

• This will probably be less like a graph and more like a family of graphs of similar functions. Pick several examples of $a$ that help you demonstrate its various characteristics. In particular, some choices of $a$ will yield a graph with $2$ $x$-intecepts, some will yield a graph with $0$ $x$-intercepts, and one will yield a graph with $1$ $x$-intedcept. All other behavior will be fairly similar, as peterwhy's answer demonstrates. – Cameron Buie Aug 13 '15 at 0:03

Setting $f(x) = 0$, \begin{align*} 1+\frac ax + \frac a{x^2} &= 0\\ x^2+ax+a &= 0\\ x &= \frac{-a\pm\sqrt{a^2-4a}}{2} \end{align*} So the $x$-intercepts, if any, depend on $a$.

Differentiating $f(x)$ w.r.t. $x$, $$f'(x) = -\frac a{x^2}-\frac {2a}{x^3} = -\frac{ax(x+2)}{x^4}$$ Setting $f'(x) = 0$, \begin{align*} -\frac a{x^2}-\frac {2a}{x^3} &= 0\\ -ax-2a&= 0\\ x&=-2 \end{align*} The $x$-coordinate of the stationary point does not depend on $a$, but the $y$-coordinate does.

Differentiating $f'(x)$ w.r.t. $x$, $$f''(x) = \frac{2a}{x^3} + \frac{6a}{x^4} = \frac{2a(x+3)}{x^4}$$ Setting $f''(x) = 0$, \begin{align*} \frac{2a}{x^3} + \frac{6a}{x^4} &= 0\\ 2ax + 6a &= 0\\ x&= -3 \end{align*} The $x$-coordinate of the inflexion point does not depend on $a$, but the $y$-coordinate does.

Consider the signs in terms of $x$, $$\begin{array}{r|c|c|c|c|c|c} x&(-\infty,-3)&-3&(-3,-2)&-2&(-2,0)&(0,\infty)\\\hline f(x)&&1-\frac a3+\frac a9&&1-\frac a2+\frac a4&&+\\\hline f'(x)&-&-&-&0&+&-\\\hline f''(x)&-&0&+&+&+&+ \end{array}$$

And also the horizontal asymptote. $$\lim_{x\to \infty}\left(1+\frac ax+\frac a{x^2}\right) = \lim_{x\to\infty} 1 + \lim_{x\to\infty} \frac ax+ \lim_{x\to\infty}\frac a{x^2} = 1$$

• Concave up is $(-3,0)\cup(0,\infty)$, because $f(0)$ is not even a thing. – peterwhy Aug 13 '15 at 18:40
• could it be, that my graph on the interval (-inf, 0) will do lower than the horizontal asymptote. I mean, not the whole graph, but its part? – Sarah Aug 13 '15 at 18:46
• yeah, I've realized about the concave. and deleted the question :) – Sarah Aug 13 '15 at 18:47
• A part of $y = f(x)$ is below $y=1$, that is when $x\in(-\infty, -a)$. – peterwhy Aug 13 '15 at 18:52

To find critical points, inflection points, etc., just do what you do when you don't have any parameters. The only difference is that the points will involve the value of $a$.

For instance, $$f'(x) = \frac{-a}{x^2} + \frac{-2a}{x^3} = \frac{-ax-2a}{x^3}.$$ Now, for which values of $x$ is this expression zero? Etc.