What is the period and phase shift of $f(x)=3\sin 4\left(x+\frac{\pi}{4}\right)-1$?

For the function what is the period and the phase shift respectively are:?

$$f(x) = 3\sin 4\left(x +\frac{\pi}{4}\right) - 1$$

My attempt: It is wrong though.

$y = a\sin b(x\pm h) \pm k$

$a$ : amplitude $\implies$ take absolute value

$b$ : angular frequency

$h$ : horizontal shift (phase shift). ($+$ left, $-$ right)

$k$ : mid. line or vertical shift ($+$ up, $-$ down)

period = $\dfrac{2\pi}{b}$ ( for sines, cosines; if tangent $p = \dfrac{\pi}{b}$)

$f(x) = 3\sin 4\left(x + \dfrac{π}{4}\right) - 1$

$b = 4$

period = $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$

phase shift: left $\dfrac{\pi}{4}$

1 Answer

I'm not sure why you say that your attempt is wrong. That's how I learned it.

Edit: As you've mentioned in the comments below, the options provided to you are

(A) period $4$ and phase shift $-\cfrac\pi4$

(B) period $\cfrac\pi2$ and phase shift $-\cfrac\pi4$

(C) period $-\cfrac\pi2$ and phase shift $-\cfrac\pi4$

(D) period $-\cfrac\pi2$ and phase shift $\cfrac\pi4$

Using the method that you were given to find the period, the only possible option is (B). Remember that a phase shift of $-\cfrac\pi4$ is the same thing as a phase shift of $\cfrac\pi4$ to the left, so this is the same as your answer.

Edit: Let me give you some alternate (hopefully clearer) rules. Suppose you are given an equation $$y=a\sin\bigl(b(x-h)\bigr)+k$$ or $$y=a\cos\bigl(b(x-h)\bigr)+k$$ for some real $a,b,h,k$ where $a\ne0$ and $b>0$. (If $a=0$ or $b=0,$ there's really nothing interesting to say. Do you know why we may assume that $b$ isn't negative?) Then:

The amplitude is $|a|$. Graphically, this will be the vertical distance from the max (or min) value to the midline, or half the vertical distance from the max value to the min value.

The phase shift is $h$. Graphically, this is a shift by $|h|$ to the right if $h$ is positive; by $|h|$ to the left if $h$ is negative.

The midline shift is $k$. Graphically, this is a shift by $|k|$ upward if $k$ is positive; by $|k|$ downward if $k$ is negative.