Is f(x) = (x+1)/(x+2) considered a hyperbola? How does it's derivate f'(x) = 1/(x+2)^2 show that it is a hyperbola? I thought that the equation of a hyperbola is in the form x^2/a^2-y^2/b^2=1. Can someone explain this concept to me? Thank you!

Question

stacey · Answer

Since it can be put into the form (x-h)(y-k)=m, it is a rectangular hyperbola, meaning its asymptotes are parallel to the x and y-axes.
(x+2)(y-1) = -1

stacey · Answer

For a rectangular hyperbola, the slope would be undefined (vertical) at x=h. As the graph approaches positive or negative infinity, the slope will approach being horizontal or equal to zero.
The derivative will be in the form f'(x) = -m/(x-h)^2

anonymous · Answer

If you take x^2/a^2-y^2/b^2=1, set a=b and rotate everything clockwise by 45 degrees, you will get xy=-a. Set a=1, you get xy=-1 or y= -1/x. Now shift it to the left by 2 to get y=-1/(x+2), shift it up by 1 you get y=1-1/(x+2) which is the same as y=(x+1)/(x+2). So clearly what you have is a hyperbola.

But how do you tell that from the derivative? That's where Stacey's answer comes into play. If the derivative has the form -m/(x-h)^2, you know that this is a hyperbola. It's not in the form x^2/a^2-y^2/b^2=1, but a 45 degree rotated version of it.