1. ## Turning points

Hi

For y=2/(x-1), how do I exactly find the turning pts?

I tried following my teacher's way by first turning it into a quadratic equation and finding its discriminant:

y(x-1) = 2
yx - y - 2 = 0
discriminant = y^2
For x to be real then its discriminant must be greater or equal to 0
i.e. y^2 is greater than or equal to 0
This becomes y is greater than or equal to 0.
When y = 0, equation becomes 0 = 2.
That's where I'm stuck. For the other graphs, my teacher's method allowed me to find the turning points. For this one (the first question ), it doesn't work.

2. Originally Posted by xwrathbringerx
Hi

For y=2/(x-1), how do I exactly find the turning pts?

I tried following my teacher's way by first turning it into a quadratic equation and finding its discriminant:

y(x-1) = 2
yx - y - 2 = 0
discriminant = y^2
For x to be real then its discriminant must be greater or equal to 0
i.e. y^2 is greater than or equal to 0
This becomes y is greater than or equal to 0.
When y = 0, equation becomes 0 = 2.
That's where I'm stuck. For the other graphs, my teacher's method allowed me to find the turning points. For this one (the first question ), it doesn't work.

$y = \frac{2}{x-1}$ does not have any turning points. It's a rectangular hyperbola.
3. But $yx - y - 2 = 0$ is not a quadratic. What 'discriminant' are you talking about then?
To get turning points (if any), find its derivative and equate it to $0$.