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?

  • $\begingroup$ 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. $\endgroup$ – 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. $$\lim_{x\to \infty}\left(1+\frac ax+\frac a{x^2}\right) = \lim_{x\to\infty} 1 + \lim_{x\to\infty} \frac ax+ \lim_{x\to\infty}\frac a{x^2} = 1$$

  • $\begingroup$ Concave up is $(-3,0)\cup(0,\infty)$, because $f(0)$ is not even a thing. $\endgroup$ – peterwhy Aug 13 '15 at 18:40
  • $\begingroup$ 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? $\endgroup$ – Sarah Aug 13 '15 at 18:46
  • $\begingroup$ yeah, I've realized about the concave. and deleted the question :) $\endgroup$ – Sarah Aug 13 '15 at 18:47
  • $\begingroup$ A part of $y = f(x)$ is below $y=1$, that is when $x\in(-\infty, -a)$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.