The graphs are shown below. The graph of [latex]g(x)= \sin (x - \dfrac{\pi}{4})[/latex] has the same amplitude, midline, and period as the graph of [latex]f(x)=\sin x{,}[/latex] but the graph of [latex]g[/latex] is shifted to the \right by [latex]\dfrac{\pi}{4}[/latex] units, compared to the graph of [latex]f{.}[/latex]
We can see why this shift occurs by studying a table of values for the two functions.

Notice that in the table, [latex]g[/latex] has the same function values as [latex]f{,}[/latex] but each one is shifted [latex]\dfrac{\pi}{4}[/latex] units to the right. The same thing happens in the graph: each [latex]y[/latex]-value appears [latex]\dfrac{\pi}{4}[/latex] units farther to the right on [latex]g[/latex] than it does on [latex]f{.}[/latex]

First notice that we can write the equation as

[latex]y=-2\cos \left[x-\left(\dfrac{-\pi}{3}\right)\right][/latex]

so that [latex]h = -\dfrac{\pi}{3}{,}[/latex] and we expect the graph to be shifted [latex]\dfrac{\pi}{3}[/latex] units to the \left compared to the graph of [latex]y = \cos x{.}[/latex]

We choose convenient values for the inputs of the cosine function; namely, [latex]x+\dfrac{\pi}{3}{.}[/latex]

From those values, we work backward to [latex]x[/latex] and forwards to [latex]y{.}[/latex] (To obtain the values of [latex]x{,}[/latex] we subtract [latex]\dfrac{\pi}{3}[/latex]from the values of [latex]x+\dfrac{\pi}{3}{.}[/latex])

To make the graph, we’ll scale the [latex]x[/latex]-axis in multiples of [latex]\dfrac{\pi}{6}{.}[/latex] We plot the guide points [latex](x,y)[/latex] from the table, and sketch a sinusoidal graph through the points. One cycle of the graph is shown below. As we expected, the graph is shifted [latex]\dfrac{\pi}{3}[/latex] units to the left compared to the graph of [latex]y = \cos x{.}[/latex]

1. First, we write the input for the cosine function in factored form:[latex]f(x) = 4\cos 3\left(x+\dfrac{\pi}{3}\right) - 2[/latex] Comparing the formula for [latex]f[/latex] with the standard form, we see that [latex]A=4{,}[/latex] [latex]B=3,~h = \dfrac{-\pi}{3}{,}[/latex] and [latex]k = 2{.}[/latex] Thus, the midline is [latex]y = -2{,}[/latex] the amplitude is [latex]4{,}[/latex] the period is [latex]\dfrac{2\pi}{3}{,}[/latex] and the horizontal shift is [latex]\dfrac{\pi}{3}[/latex] units to the left.
2. Use a table of values to locate the guide points for the graph. Begin by choosing convenient values for the input, [latex]3x+\pi{.}[/latex] Then work backward to find values for [latex]x[/latex] and forward to find values for [latex]f(x){.}[/latex]
   

   [latex]x[/latex]	[latex]3x[/latex]	[latex]3x+\pi[/latex]	[latex]\cos(3x+\pi)[/latex]	[latex]4\cos 3(x+\dfrac{\pi}{3})-2[/latex]
   [latex]\dfrac{-\pi}{3}[/latex]	[latex]-\pi[/latex]	[latex]0[/latex]	[latex]1[/latex]	[latex]2[/latex]
   [latex]\dfrac{-\pi}{6}[/latex]	[latex]\dfrac{-\pi}{2}[/latex]	[latex]\dfrac{\pi}{2}[/latex]	[latex]0[/latex]	[latex]-2[/latex]
   [latex]0[/latex]	[latex]0[/latex]	[latex]\pi[/latex]	[latex]-1[/latex]	[latex]-6[/latex]
   [latex]\dfrac{\pi}{6}[/latex]	[latex]\dfrac{\pi}{2}[/latex]	[latex]\dfrac{3\pi}{2}[/latex]	[latex]0[/latex]	[latex]-2[/latex]
   [latex]\dfrac{\pi}{3}[/latex]	[latex]\pi[/latex]	[latex]2\pi[/latex]	[latex]1[/latex]	[latex]2[/latex]

   Finally, plot the guide points and connect them with a sinusoidal curve. The graph is shown at right.
   

The midline of the graph is [latex]y=0{,}[/latex] and its amplitude is [latex]4{.}[/latex] It has the shape of a cosine, but its maximum value occurs at [latex]x=\dfrac{\pi}{6}[/latex] instead of at [latex]x=0{,}[/latex] so the graph is shifted [latex]\dfrac{\pi}{6}[/latex] units to the right compared to the graph of [latex]y=\cos x{.}[/latex]

The graph completes one cycle between [latex]x=\dfrac{\pi}{6}[/latex] and [latex]x=\dfrac{13\pi}{6}{,}[/latex] so its period is [latex]2\pi{.}[/latex] Thus, we use the standard form [latex]y = A\cos B(x-h) + k[/latex] with [latex]A=4,~B=1,~h=\dfrac{\pi}{6}{,}[/latex] and [latex]k=0{,}[/latex] giving us

[latex]y = 4\cos (x-\dfrac{\pi}{6})[/latex]

as our formula for the function.

1. The graph has the shape of [latex]y=-\cos x{.}[/latex] It appears to have midline [latex]y=75[/latex] and amplitude [latex]75{,}[/latex] so [latex]k=75[/latex] and [latex]A=-75[/latex]. To find [latex]B{,}[/latex] we compute
   [latex]B=\dfrac{2\pi}{{period}}=\dfrac{2\pi}{10.8}=0.58[/latex]
   Thus, a formula for the model is [latex]y=75-75\cos[0.58(x-h)]{.}[/latex]
2. Fitting a curve by eye is a subjective process, but it looks like we can get a better fit by shifting the curve slightly to the left. By trying several values for [latex]h[/latex] in [latex]y=75-75\cos[0.58(x-h)]{,}[/latex] we might settle on [latex]h=0.3{,}[/latex] which results in the curve shown below.