Ples help me find slant assemtotesthis one is different because it isn't a rational functionfind the slant assemtotes of [tex](y+1)^2=4xy[/tex]the equation can be rewritten using the quadratic formula as [tex]y=2x-1 \pm \sqrt{x^2-x}[/tex]ples find slant assemtotes and show all workthx

A polynomial asymptote is a function [tex]p(x)[/tex] such that

[tex]\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0[/tex]

[tex](y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}[/tex]

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form [tex]p(x)=ax+b[/tex].

Ignore the negative root (we don't need it). If [tex]y=2x-1+2\sqrt{x^2-x}[/tex], then we want to find constants [tex]a,b[/tex] such that

[tex]\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0[/tex]

We have

[tex]\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}[/tex]
[tex]\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}[/tex]
[tex]\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}[/tex]

since [tex]x\to\infty[/tex] forces us to have [tex]x>0[/tex]. And as [tex]x\to\infty[/tex], the [tex]\dfrac1x[/tex] term is "negligible", so really [tex]\sqrt{x^2-x}\approx x[/tex]. We can then treat the limand like

[tex]2x-1+2x-ax-b=(4-a)x-(b+1)[/tex]

which tells us that we would choose [tex]a=4[/tex]. You might be tempted to think [tex]b=-1[/tex], but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of [tex]b[/tex], we have to solve for it in the following limit.

[tex]\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0[/tex]

[tex]\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2[/tex]

We write

[tex](\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}[/tex]

Now as [tex]x\to\infty[/tex], we see this expression approaching [tex]-\dfrac12[/tex], so that

[tex]-\dfrac12=\dfrac{b+1}2\implies b=-2[/tex]

So one asymptote of the hyperbola is the line [tex]y=4x-2[/tex].

The other asymptote is obtained similarly by examining the limit as [tex]x\to-\infty[/tex].

[tex]\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0[/tex]

[tex]\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0[/tex]

Reduce the "negligible" term to get

[tex]\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0[/tex]

Now we take [tex]a=0[/tex], and again we're careful to not pick [tex]b=-1[/tex].

[tex]\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0[/tex]

[tex]\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2[/tex]

[tex](x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}[/tex]

This time the limit is [tex]\dfrac12[/tex], so

[tex]\dfrac12=\dfrac{b+1}2\implies b=0[/tex]

which means the other asymptote is the line [tex]y=0[/tex].