Ytukyg

Why can’t you create an equation of a line that gets closer to a line as it heads to $infty,$ such as $y=x$ or $y=2x+4$ and a horizontal asymptote that approaches something like $-2$ as it goes towards $-infty.$ Can somebody please help me build an equation for this line. (https://i.stack.imgur.com/kYaVS.jpg)

As said by @Torsten Schoeneberg there are many ways to obtain such curves. However, (branches of) hyperbolas rank among the most natural ones.

Thus I propose the following general solution among the hyperbola family : for any real parameter $a>0$

$$y=dfrac12left(x-2+sqrt{(x+2)^2+4a}right)$$

Fig. Different branches for $a=0.5, 1, 1.5, ...4$.

How did I find this general equation ?

I have first considered the curve reduced to its asymptotes, i.e., with equation

$$(y+2)(y-x)=0 tag{1}$$

(the product of equations $y+2=0$ and $y-x=0$). Then I pertubated a little this equation by taking instead of it :

$$(y+2)(y-x)-a=0 tag{2}$$

Now expand it as a quadratic in variable $y$ :

$$y^2+y(2-x)-(2x+a)=0 tag{3}$$

and use the classical formula for the roots of a quadratic considering $x$ as a parameter.

An example of a function with asymptotes $;y=-2;$ as $;xto -infty;$ and $;y=x-2;$ as $xto infty;$ is $$f(x)=frac{sqrt{x^2+1}+x}{2}-2$$

To amend my second comment, for what it's worth, here is my thought process how to get a function as demanded, and what I experienced when doing that "shifting and stretching".

So to start, I looked for a function with two different horizontal asymptotes for $x to -infty$ versus $x to infty$. Well, the $arctan$ came to my mind. Now first I wanted to turn that in the general shape "0 on the left, but positive (so I can make it increasing) on the right". So I moved it up by $pi/2$, whence it would go from $0$ at $-infty$ to $pi$ (or whatever positive value) at $infty$.

$arctan(x)+frac{pi}{2}$:

This, multiplied with $frac{x}{pi}$, will give something that will behave like $0$ to the left, but like $x$ on the right (so I thought):

$displaystyle frac{x}{pi} cdot left(arctan(x) + frac{pi}{2}right)$:

Now shift it down by $2$. That will make $lim_{xto -infty} =-2$, I thought. It also moves the slant asymptote down to $y=x-2$. To make good for that, I thought I move the graph to the left by $2$, which transforms the slant asymptote back to $y=x$ and does not change the horizontal one.

$displaystyle frac{x+2}{pi} cdot left(arctan(x) + frac{pi}{2}right) -2$:

But now I noticed that ever since the second step I was off by a vertical shift of $1/pi$. See, both asymptotes are a bit too low. Where does that come from? Well I realised that even though arctan goes to $pmpi/2$ (and my shifted version in the first step to $0$ resp. $pi$), I multiply it with a linear term; which means that I have to find out how well the $arctan$ approaches its limit up to a term of order $frac1x$. And indeed, an expansion of $arctan$ "around $pm infty$" is given by

$$displaystyle arctan(x) = frac{pi}{2} - frac{1}{x} + frac{1}{3x^{3}} - frac{1}{5x^{5}} + ...$$ (for $x >1$) resp.

$$displaystyle arctan(x) = -frac{pi}{2} - frac{1}{x} + frac{1}{3x^{3}} - frac{1}{5x^{5}} + ...$$ (for $x < -1$).

whose $-1/x$-term, by multiplication with $x/pi$, indeed explained that constant $-1/pi$ term. But then that's that, one just removes it by hand, giving:

$$f(x) = displaystyle frac{x+2}{pi} cdot left(arctan(x) + frac{pi}{2}right) -2 +frac{1}{pi}$$

Voilà:

Fun fact I learned on the way when deriving the "expansion around $infty$": the derivative of $arctan(1/x)$ is $(-1)cdot$ the derivative of $arctan(x)$; which abstractly implies that $arctan(x)= C - arctan(1/x)$ for some constant depending on the connected component of the graph we are on. And indeed e.g. for $x>0$, we have $arctan(x) = pi/2 - arctan(1/x)$.