# How would I find the vertical and horizontal asymptotes of a y=1/x function algebraically? For example, y=−2/(x+3) −1 (as you would type into a calculator). Simply, how do I find the x and y values by looking at this equation? In other words, the middle point where the asymptotes in the picture has moved and the whole graph has been vertically stretched, where are the asymptotes now?

Identifying Asymptotes of a Hyperbola
basic hyperbolic functionHow would I find the vertical and horizontal asymptotes of a $y=\frac{1}{x}$ function algebraically? For example, $y=-\frac{2}{x+3}-1$ (as you would type into a calculator). Simply, how do I find the x and y values by looking at this equation? In other words, the middle point where the asymptotes in the picture has moved and the whole graph has been vertically stretched, where are the asymptotes now?
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SoroAlcommai9
I assume that's $y=-\frac{2}{x+3}-1$. There's a vertical asymptote where the function is undefined, and the function is undefined where it involves division by zero. For the horizontal asymptote, you have to ask yourself, what happens when x grows without bound? when x gets very large negative?
EDIT: Reading OP's comments and revised question, I think maybe this approach will be more what's wanted.
I take it you know that for y=1/x the horizontal asymptote is y=0 and the vertical asymptote is x=0, and they meet at the point (0,0). y=1/(x+3) is the same graph but shifted 3 units to the left; that doesn't change the horizontal asymptote, but it shifts the vertical asymptote to x=−3. y=−1/(x+3) flips (or, reflects) the graph in the x-axis, but has no effect on the asymptotes, which remain y=0 and x=−3. y=−2/(x+3) stretches the graph, but again has no effect on the asymptotes. Finally, $y=\frac{-2}{x+3}-1$ lowers the graph by 1, so it changes the horizontal asymptote to y=−1, while not affecting the vertical asymptote.
So, the new asymptotes are y=−1 and x=−3, which meet at the point (−3,−1), which is what you are calling the middle point.