I tried to follow the rules of sketching a graph, I found the intersection with the ox and oy in (0,1)and (-1,0) and I think that the oblique asymptote is y=x+4. What about the vertical ones? The roots of the denominator are complex ones. Can we say these are asymptotes?.. The first derivative is pretty hard to find roots of. How can I sketch its graph, please?
Sketch the graph of the function $f(x) =\frac{(x+1)^3} {x^2 – x+1}$
calculusgraphing-functions
Related Solutions
It's not correct to say that the x-intercept is at $x=-1$. The x-intercept occurs when $f(x) = 0$, so we have to solve the following:
$$\begin{align} \ln x + 2 &= 0 \\ \ln x &= -2 \\ x &= e^{-2} \approx 0.13\end{align} $$
While your y-intercept is sort of correct, it doesn't make a lot of sense to say the the regular $\ln x $ funcion has a y-intercept at $y=-\infty$. Rather, you should say that the y-intercept is the value of $f(x)$ at $x=0$. But $f(x) = \ln x + 2$ is not defined at $x=0$, so this funcion doesn't have a y-intercept.
You mention at the end that you know this funcion is like $\ln x$ moved up two units. That should be all you need to know: It will still have the vertical asymptote at $x=0$, it will still go to infinity as $x$ goes to infinity, it will still have only one x-intercept. The shape is the same, but moved vertically.
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. $$\lim_{x\to \infty}\left(1+\frac ax+\frac a{x^2}\right) = \lim_{x\to\infty} 1 + \lim_{x\to\infty} \frac ax+ \lim_{x\to\infty}\frac a{x^2} = 1$$
Best Answer
You have $$f(x) =\frac{(x+1)^3} {x^2 - x+1} = x + 4 + \frac{6x - 3}{x^2-x+1}.$$
As $\lim\limits_{x \to \pm \infty} \frac{6x - 3}{x^2-x+1} = 0$, $y=x+4$ is indeed an asymptote of $f$. Also the sign of $\frac{6x - 3}{x^2-x+1}$ is the one of $6x-3$ as $x^2-x+1$ is always positive. So $f$ crosses its asymptote at $x = 1/2$, is above it for $x > 1/2$ and below it for $x < 1/2$.
You can also find where the maximal vertical distance between $f$ and its asymptote is by vanishing the derivative of $g(x)=\frac{6x - 3}{x^2-x+1}$. You'll find two symmetric values about $x=1/2$.
Some elements to start a decent plot...