# Turning points

• Nov 16th 2008, 12:41 AM
xwrathbringerx
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 (Crying)), it doesn't work.

• Nov 16th 2008, 12:43 AM
mr fantastic
Quote:

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 (Crying)), it doesn't work.

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