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Let $f$ be a generic rational function such that

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

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

Desmos : https://www.desmos.com/calculator/y9eue8tdr5

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.

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    $\begingroup$ 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. $\endgroup$ – Andrew Chin Mar 10 at 18:37
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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.

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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.

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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 $\sqrt a$.

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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 $a_1<a_2$, the Graph of $f_{a_1}$ snuggles the asymptote in a faster way than the Graph of $f_{a_2}$.
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