Graphing the trig function from to given points?

**SnowBudgie** · March 18th 2011, 09:56 PM

Leave the answer in radical form unless told otherwise. It is easier to write and remember.

**Prove It** · March 18th 2011, 09:57 PM

$\displaystyle -4 = \frac{a}{\sqrt{2}}$ .

How do you undo dividing by $\displaystyle \sqrt{2}$ ?

**Smokinoakum** · March 18th 2011, 10:08 PM

Originally Posted by Prove It

$\displaystyle -4 = \frac{a}{\sqrt{2}}$ .

How do you undo dividing by $\displaystyle \sqrt{2}$ ?

I guess I'm more of a visual learner, which is why I need to see the full problem worked out, study the process, understand the logic, and then apply what I learn.

So (a = 4)

I still would like to see the whole problem worked out for my notes. I still confused.

**skeeter** · March 19th 2011, 06:36 AM

note that $a \ne 4$

$-4 = \dfrac{a}{\sqrt{2}}$

to solve for $a$ , multiply both sides by $\sqrt{2}$ ...

$a = -4\sqrt{2}$

also, your instructor should have introduced you to the exact trig values of sine and cosine on the unit circle. (then you would have known the exact value of $\sin\left(\frac{3\pi}{4}\right)$ w/o consulting your calculator)

**Smokinoakum** · March 19th 2011, 08:13 AM

Originally Posted by skeeter

note that $a \ne 4$

$-4 = \dfrac{a}{\sqrt{2}}$

to solve for $a$ , multiply both sides by $\sqrt{2}$ ...

$a = -4\sqrt{2}$

also, your instructor should have introduced you to the exact trig values of sine and cosine on the unit circle. (then you would have known the exact value of $\sin\left(\frac{3\pi}{4}\right)$ w/o consulting your calculator)

Very cool! That helps a lot.

Well I entered in $a = -4\sqrt{2}$ in my assignment and it said that it was wrong?

The answer was (the function y = ( 4 ) sin x passes through (3pi/2 , -4)

I think I've confused both myself and you generous people helping me out. Such is learning this stuff online and not in a class room.

What have I missed here?

**Smokinoakum** · March 19th 2011, 08:44 AM

Originally Posted by Prove It

Like I said, substitute $\displaystyle x = \frac{3\pi}{4}$ and $\displaystyle y = -4$ into the equation $\displaystyle y = a\sin{x}$ .

Surely you can go from there...

This is where some of the confusion happened!

He should have said "substitute $\displaystyle x = \frac{3\pi}{2}$ and $\displaystyle y = -4$ into the equation"

Not "substitute $\displaystyle x = \frac{3\pi}{4}$ and $\displaystyle y = -4$ into the equation"

So the equation should be $\displaystyle -4 = a\ sin \frac{3\pi}{2}$ right?

**Smokinoakum** · March 19th 2011, 09:45 AM

OK...going from what Prove It helped me out with, this is what I have figured out.

Find the function of the form $\displaystyle y = a\ sin\ x$ that passes through $\displaystyle \left(\frac{3\pi}{2}, -4\right)$

Substitute $\displaystyle (x, y) = \left(\frac{3\pi}{2}, -4\right)$ into $\displaystyle y=a\sin{x}$ and solve for $\displaystyle a$ .

$\displaystyle -4 = a\ sin \frac{3\pi}{2}$

$\displaystyle sin \frac{3\pi}{2} =\ -1$

$\displaystyle -4 =\frac{a}{-1}$

$\displaystyle -4 (-1) =\ 4$

$\displaystyle a =\ 4$

Does this follow the concept?

**Prove It** · March 19th 2011, 04:39 PM

Originally Posted by Smokinoakum

OK...going from what Prove It helped me out with, this is what I have figured out.

Find the function of the form $\displaystyle y = a\ sin\ x$ that passes through $\displaystyle \left(\frac{3\pi}{2}, -4\right)$

Substitute $\displaystyle (x, y) = \left(\frac{3\pi}{2}, -4\right)$ into $\displaystyle y=a\sin{x}$ and solve for $\displaystyle a$ .

$\displaystyle -4 = a\ sin \frac{3\pi}{2}$

$\displaystyle sin \frac{3\pi}{2} =\ -1$

$\displaystyle -4 =\frac{a}{-1}$

$\displaystyle -4 (-1) =\ 4$

$\displaystyle a =\ 4$

Does this follow the concept?

Sorry about the confusion before, thought it said $\displaystyle x = \frac{3\pi}{4}$ .

Yes, you have now correctly found $\displaystyle a$ .

So the equation of your function is $\displaystyle y = 4\sin{x}$ . This is the same as the function $\displaystyle \sin{x}$ , being stretched by a factor of $\displaystyle 4$ vertically.

**Smokinoakum** · March 19th 2011, 04:55 PM

No worries about the confusion. It made me really try to fully grasp how these functions relate to one another. Thanks a lot for your help.

It's great to have a place like this to come to when you take math courses online!

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