How do you graph #y=cotx+2#?

Question

Daniel Acosta · Accepted Answer

graph{cotx+2 [-10, 10, -5, 5]}

Yeah, that's about it.

What you do is prepare a table of values with an appropriate range like, #-5<=x<=5# and plot on a graph

Daniel Acosta · Answer

undefined To graph $ y = \cot(x) + 2 $, follow these steps:

1. Determine the asymptotes: 
   - Vertical asymptotes occur where the denominator of the cotangent function, which is $ \sin(x) $, equals zero. So, the vertical asymptotes occur at $ x = k\pi $, where $ k $ is an integer.
   
2. Plot the intercept:
   - The function intersects the y-axis when $ x = 0 $. So, when $ x = 0 $, $ y = \cot(0) + 2 = \frac{1}{\tan(0)} + 2 = \frac{1}{0} + 2 = \infty + 2 = \text{undefined} $. There's no y-intercept.
   
3. Determine the behavior near the asymptotes:
   - As $ x $ approaches $ k\pi $ from the left, $ \cot(x) $ approaches $ -\infty $, and as $ x $ approaches $ k\pi $ from the right, $ \cot(x) $ approaches $ +\infty $. Adding 2 shifts this behavior upwards by 2 units.

4. Plot additional points and sketch the graph:
   - You can plot additional points by selecting different values of $ x $ and calculating the corresponding $ y $ values using the equation $ y = \cot(x) + 2 $. Then, sketch the graph based on the behavior near the asymptotes and the plotted points.

5. Draw the graph:
   - Connect the points smoothly, maintaining the behavior around the asymptotes, to complete the graph of $ y = \cot(x) + 2 $.

Carter Anderson · Answer

undefined To graph $ y = \cot(x) + 2 $, you can start by plotting the parent function $ y = \cot(x) $, which is a periodic function with vertical asymptotes at $ x = k\pi $ for integer values of $ k $, and horizontal asymptotes at $ y = 0 $. Then, shift the entire graph vertically upwards by 2 units to obtain the graph of $ y = \cot(x) + 2 $.