The directrix and focus of a parabola are shown on the graph. Which is an equation for the parabola?
2 answers:
You might be interested in
The function
... y = 1/x
has derivative
... y' = -1/x²
which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.
Your function
... y = (x+1)/(x-3)
can be written as
... y = 1 +4/(x-3)
which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).
The critical point is x=3.
The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.
