## chaotic_butterflies one year ago Multiple choice - What sine function represents an amplitude of 4, a period of pi/2, no horizontal shift, and a vertical shift of −3?

1. chaotic_butterflies

a)f(x) = -3 sin 4x +4 b) f(x)= 4 sin 4x -3 c) f(x) = 4 sin (pi/2) x -3 d) f(x) = -3 sin (pi/2) x + 4

2. anonymous

For the function $\large a ~sin (bx-c)+d$ $\large amplitude= \left| a \right|$$\large period=\frac{2\pi}{b}$$\large vertical ~shift = d$

3. chaotic_butterflies

I'm not exactly sure how to plug in the inside of the parentheses. So far I have $\left| 4 \right| \sin (???) - 3$ @LegendarySadist

4. anonymous

You can disregard the c in this example, as it's used for the phase shift, aka horizontal shift. So the b will be the integer before the $$x$$

5. chaotic_butterflies

I knew that c was the horizontal shift, but what exactly is b?

6. anonymous

b is just the number before the x. For option #2, b is 4 and for option #3 b is $$\large \frac{\pi}{2}$$

7. chaotic_butterflies

I can see that b is before x, but I don't understand what I should plug in for b, or how to figure that out.

8. anonymous

So we are looking for b, and we know the period. We'd set it up like this. $\large \frac{2\pi}{b}=\frac{\pi}{2}$ Now cross multiply and solve for b

9. chaotic_butterflies

so b = 2?

10. anonymous

$\large \frac{2\pi}{2} = \frac{\pi}{2}\\\large \pi = \frac{\pi}{2}?$ No b is not 2

11. chaotic_butterflies

Herp, I forgot to multiply 2pi by pi!

12. anonymous

$\large \frac{2\pi}{b} = \frac{\pi}{2}$ Cross multiplying will get you $\large 4\pi=\pi b$

13. chaotic_butterflies

Can't you then just cancel out the pi's and get 4 = b?

14. anonymous

Yep, that would be it. Good job :)

15. chaotic_butterflies

Thank you very much!

16. anonymous

$\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile$

17. chaotic_butterflies

c: