How do you find out the gradient of the equation of the line that is the derivative of the turning point of a hyperbola? A simple hyperbola has the gradient of 1 or -1 for the gradient of the line that is the equation of the derivative at the turning points of it(depending on whether it's in the second and fourth or the first and third quadrants) but I'm not sure how to do it for a more complicated one, the ones that have asymptotes that are's straight up and down or straight across. If someone tells me or I figure it out on my own I can do my maths exam much quicker, no one else seems to be trying to find out how to get the turning points of hyperbolas, sketching is so much quicker if you can do it. I don't even know if they'll have more complex hyperbolas in the test, but you never know and it helps to be prepared. Thanks in advance.

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