# Rational function $\frac {x^2+a} {x}$ : how does number $a$ influence the graph of the function?

[EDITED]

Let $$f$$ be a generic rational function such that

$$f(x)=$$ $$\frac {x^2+a}{x}$$.

Let number $$a$$ vary from $$0$$ to $$5$$.

What I am trying to understand is the influence the value of number $$a$$ has on the graph of $$f$$.

I'm specially interested in what happens in the quadrant on the left, below the X-axis.

What I understand is that when the absolute value of $$x$$ gets large ( as $$x$$ goes to " minus infinity"), the number $$a$$ becomes insignificant, and $$f(x)$$ becomes practically equal to $$\frac{x^2} {x}$$ = $$x$$.

So, in a way, for large absolute values of $$x$$, the graph of $$f$$ is nearly the same as the straight line $$y= x$$.

However, at a certain point ( say, around $$x = -15$$ up to $$x=0$$) the graph of $$f$$ separates itself from the graph of $$y=x$$ and begins to go down.

What I do not understand is not why the graph "sinks" ( this is due to the asymptote phenomenon, as $$x$$ approaches $$0$$). My question is: what influence has the value of number $$a$$ as to the " moment" $$f(x)$$ leaves the direction $$y=x$$ ( when one looks at the graph in the indicated quadrant, from the left to the right).

Maybe, I could say that, the more $$a$$ is small, the more $$f(x)$$ is similar to $$y=x$$ ( up to the moment $$a$$ becomes $$0$$, so that $$f$$ is actually the same as $$y=x$$); and, in the opposite sense, the more $$a$$ increases, the more the two functions get dissimilar.

But I would like a more algebraic explanation.

• This could be worth a read; you have a hyperbola with the equation $x^2-xy+a=0$, and the position of the foci changes as $a$ changes. – Andrew Chin Mar 10 at 18:37

Note that $$f(x)=x+\frac {a}{x}$$

Thus for positive values of $$a$$ you have a vertical asymptote at $$x=0$$ and a slant asymptote of $$y=x$$

The turning points of your graph are a local minimum at $$(\sqrt a, 2\sqrt a)$$ and a local maximum at $$(-\sqrt a, -2\sqrt a)$$.

The shape of the graph does not change much with variation of $$a$$ as long as $$a$$ is positive.

Well, the the role of a becomes more apparent when you consider the derivative of f. It's simply that. The roots of the equation you get when you differentiate f will give the "moment" at which it leaves the line y=x. Also, as your question is quite unclear, I want to add that since the function is not a continuous one, you should consider the different branches on the 2 sides of the vertical asymptote separately.

I'm assuming that $$a$$ is positive. Consider the function $$g(x)=\dfrac{x^2+1}{x}$$. Then $$\sqrt{a}\cdot g\left(\frac{x}{\sqrt a}\right)=\dfrac{x^2+a}{x}$$. Hence, the graph of $$f$$ looks similar to that of $$g$$ and can be obtained from the graph of $$g$$ by scaling: first, along the $$x$$-axis by a factor of $$\sqrt a$$, then along the $$y$$-axis by a factor of $$\sqrt a$$.

That's not complicated! You can explain all that easier by writing $$f_a(x)=\frac{x^2}{x}+\frac{a}{x}$$

• If $$a=0$$, that is $$f(x)=x$$ if $$x\neq 0$$ and you can make this continuous with $$f(0)=0$$.
• If $$a\neq 0$$, choose two values of $$a$$, namely $$a_1,a_2$$, and compare $$f_{a_1}$$ with $$f_{a_2}$$. If $$a_1, the Graph of $$f_{a_1}$$ snuggles the asymptote in a faster way than the Graph of $$f_{a_2}$$.