Question

find an equation for a negative cosine function having an amplitude of 3, a period of $\pi\sqrt{2}$, and a horizontal shift of $\frac{\pi\sqrt{2}}{12}$ to the left.
$y = ? \cos(\sqrt{}x + \frac{\pi}{})$

Explanation:

Step1: Recall the general form of a cosine function

The general form of a cosine function is \( y = A\cos(Bx + C) + D \). For a negative cosine function, \( A \) will be negative. The amplitude is \( |A| \), the period is \( \frac{2\pi}{|B|} \), and the horizontal shift is \( -\frac{C}{B} \) (a shift to the left is positive \( -\frac{C}{B} \), a shift to the right is negative \( -\frac{C}{B} \)). Here, \( D = 0 \) (no vertical shift).

Step2: Determine the value of \( A \)

The amplitude is given as 3. Since it's a negative cosine function, \( A=- 3 \).

Step3: Determine the value of \( B \)

The period \( T=\pi\sqrt{2} \). Using the period formula \( T = \frac{2\pi}{|B|} \), we can solve for \( B \):

$$ \pi\sqrt{2}=\frac{2\pi}{|B|} $$

Divide both sides by \( \pi \):

$$ \sqrt{2}=\frac{2}{|B|} $$

Then, \( |B|=\frac{2}{\sqrt{2}}=\sqrt{2} \). Since there's no reflection affecting \( B \) (the negative is in \( A \)), \( B = \sqrt{2} \) (we can take \( B>0 \) as the negative is in \( A \)).

Step4: Determine the value of \( C \)

The horizontal shift is \( \frac{\pi\sqrt{2}}{12} \) to the left. The horizontal shift formula is \( \text{Shift}=-\frac{C}{B} \). We know \( \text{Shift}=\frac{\pi\sqrt{2}}{12} \) and \( B = \sqrt{2} \), so:

$$ \frac{\pi\sqrt{2}}{12}=-\frac{C}{\sqrt{2}} $$

Multiply both sides by \( \sqrt{2} \):

$$ \frac{\pi\sqrt{2}\times\sqrt{2}}{12}=-C $$

Simplify \( \sqrt{2}\times\sqrt{2} = 2 \):

$$ \frac{\pi\times2}{12}=-C $$

$$ \frac{\pi}{6}=-C $$

So, \( C = -\frac{\pi}{6} \)? Wait, no, wait. Wait, the horizontal shift for \( y = A\cos(Bx + C) \) is \( -\frac{C}{B} \). A shift to the left by \( h \) means \( -\frac{C}{B}=h \), so \( C=-Bh \). Let's re - do this.

Given horizontal shift \( h=\frac{\pi\sqrt{2}}{12} \) (left shift), and \( B = \sqrt{2} \). Then \( C=-Bh=- \sqrt{2}\times\frac{\pi\sqrt{2}}{12} \)

$$ C=-\frac{\pi\times2}{12}=-\frac{\pi}{6} $$

But in the equation \( y = A\cos(Bx + C) \), the argument is \( Bx + C=\sqrt{2}x-\frac{\pi}{6} \)? Wait, no, wait. Wait, if the shift is to the left, the formula is \( y = A\cos(B(x + h)) \), which expands to \( y = A\cos(Bx + Bh) \). Let's use this form. So \( y=A\cos(B(x + h)) \), where \( h \) is the left shift.

So \( y=-3\cos(\sqrt{2}(x+\frac{\pi\sqrt{2}}{12})) \)
Expand the argument:

$$ \sqrt{2}(x+\frac{\pi\sqrt{2}}{12})=\sqrt{2}x+\frac{\pi\times2}{12}=\sqrt{2}x+\frac{\pi}{6} $$

So the function is \( y=-3\cos(\sqrt{2}x + \frac{\pi}{6}) \)

So comparing with \( y = [?]\cos(\sqrt{[ ]}x+\frac{\pi}{[ ]}) \), we have:

• The first box (coefficient of cosine) is \( - 3 \)
• The second box (inside the square root with \( x \)) is \( 2 \) (since \( B = \sqrt{2}\), so \( B^{2}=2 \)? Wait, no, the argument is \( \sqrt{2}x+\frac{\pi}{6} \), so the square root term is \( \sqrt{2} \), so the number inside the square root is \( 2 \)
• The third box (denominator of \( \pi \)) is \( 6 \)

Answer:

The first box is \(-3\), the second box is \(2\), the third box is \(6\). So the equation is \( y=-3\cos(\sqrt{2}x + \frac{\pi}{6}) \) and filling in the boxes: \( y=\boxed{-3}\cos(\sqrt{\boxed{2}}x+\frac{\pi}{\boxed{6}}) \)