Question

How do you draw the graph of `y=2-cosx` for `0<=x<2pi`?

Answer 1

See below

Explanation:

This excercise involves function transformations. It means that you start from a function whose graph is known, and tranform it. Let's see what the transformations are and how they affect the graph:

Transformation 1: sign change

As a first step, we change from `cos(x)` to `-cos(x)`. In general, everytime you change from `f(x)` to `-f(x)` you reflect the graph vertically, with respect to the `x` axis. Here is the transformation:

Original function `f(x)=cos(x)`
graph{cos(x) [-0.2,6.48,-1.2,3.2]}

Reflected function `-f(x)=-cos(x)`
graph{-cos(x) [-0.2,6.48,-1.2,3.2]}

Transformation 2: vertical shift

The next transformation is represented by an additive constant: you change from `-cos(x)` to `-cos(x)+2`. In general, everytime you change from `f(x)` to `f(x)+k` you have a vertical translation, `k` units up if `k>0`, down otherwise. In this case, we have a translation of `2` units up. Here is the transformation:

Original function `f(x)=-cos(x)`
graph{-cos(x) [-0.2,6.48,-1.2,3.2]}

Shifted function `-f(x)=-cos(x)+2`
graph{-cos(x)+2 [-0.2,6.48,-1.2,3.2]}