Question

How do you graph `y = 3/4 cos 3 (x - 2) - 1`?

Answer 1

graph{3/4cos(3(x-2))-1 [-5, 5, -3, 3]}

Explanation:

We will construct this graph in the following sequence of steps:

1. `y=cos(x)`
   graph{cos(x) [-5, 5, -3, 3]}

2. `y=cos(3x)`
   This transformation squeezes the graph horizontally along the X-axis towards Y-axis by a factor of `3` because, if point `(a,b)` belongs to graph of `y=f(x)` (that is, `a` and `b` satisfy `b=f(a)` equation) then point `(a/K,b)` belongs to graph `y=f(Kx)` since `f(Ka/K)=f(a)=b`
   graph{cos(3(x)) [-5, 5, -3, 3]}

3. `y=3/4cos(3x)`
   This transformation stretches the graph vertically along the Y-axis by a factor of `3/4` because, if point `(a,b)` belongs to graph of `y=f(x)` (that is, `a` and `b` satisfy `b=f(a)` equation) then point `(a,Kb)` belongs to graph `y=Kf(x)` since `Kf(a)=Kb`
   graph{3/4cos(3(x)) [-5, 5, -3, 3]}

4. `y=3/4cos(3(x-2))`
   This transformation shifts the graph horizontally along the X-axis by `2` to the right because, if point `(a,b)` belongs to graph of `y=f(x)` (that is, `a` and `b` satisfy `b=f(a)` equation) then point `(a+delta,b)` belongs to graph `y=f(x-delta)` since `f(a+delta-delta)=f(a)=b`
   graph{3/4cos(3(x-2)) [-5, 5, -3, 3]}

5. `y=3/4cos(3(x-2))-1`
   This transformation shifts the graph vertically along the Y-axis by `1` down because, if point `(a,b)` belongs to graph of `y=f(x)` (that is, `a` and `b` satisfy `b=f(a)` equation) then point `(a,b-delta)` belongs to graph `y=f(x)-delta` since `f(a)-delta=b-delta`
   graph{3/4cos(3(x-2))-1 [-5, 5, -3, 3]}