# How to rewrite in the form of a basic rectangular hyperbola.

I was wondering how you would write the function $$y=\frac{4x-7}{x+1}$$ in the form of $$\frac{a}{x+1}+k$$.

I know the vertical asymptote is at $$x=-1$$.

I know that to find a (the dillation) that you usually need a point. The asymptotes are usually given in the question, but not this time. I also know how to change the form with linear and quadratic functions, but I seem to be getting stuck with these types of questions.

I tried solving algebraically, but I seem to always get stuck. I'm not sure what I should do first. I've tried plotting points but I cannot figure out what the equation would be from the graph.

When I mean algebraically, I mean that I tried bringing $$x+1$$ over to the other side, so that it would become $$y = \frac{4x-7}{x+1}+k.$$ However I do not know where I should go from here.

Any help would be appreciated. Thank you very much.

• No trick is needed. $$\frac{a}{x+1}+k=\frac{k(x+1)+a}{x+1}$$ so you just have to solve $$k=4,\quad k+a=-7.$$ Commented May 31, 2023 at 6:08
• "I tried solving algebraically, but I seem to always get stuck": Please edit your post to include your attempts. Quick beginner guide for asking a well-received question + please avoid "no clue" questions. Commented May 31, 2023 at 6:12
• @AnneBauval edited, thank you. Commented May 31, 2023 at 6:23
• Your edit ($y - k= \frac{4x-7}{x+1}$) is not consistent with the context ($y=\frac{4x-7}{x+1}=\frac{a}{x+1}+k$). Did you follow my first comment, instead? Commented May 31, 2023 at 7:13

$$y=\frac{4x-7}{x+1}=\frac{4x+4-11}{x+1}$$ $$\implies y= \frac{4(x+1)-11}{x+1}$$ $$\implies y=4-\frac{11}{x+1}$$ Here $$a=-11$$ and $$k=4$$
• @verygood101 I didn't get that, i got $$4x-7=4x+4-11$$ as we know that $7=11-4$ so $-7=4-11$ Commented May 31, 2023 at 6:27