Exponential equation. Points (-2,9) (0,1). My answer y=3(1/3)^x. Am I correct?

- anonymous

Exponential equation. Points (-2,9) (0,1). My answer y=3(1/3)^x. Am I correct?

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- anonymous

Well
\[3*(1/3)^{-2} = 3*3^2 = 9*3 = 27 \ne 9\] so nope. That can't be right.

- anonymous

hmm well lets see. First I got a=9/b^2. is that correct?

- anonymous

From the second point we have
\[1 = ab^0\]
Since
\[b^0=1\rightarrow 1 = ab^0 = a*1 \rightarrow a = 1\]

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## More answers

- anonymous

Try plotting y = 2^x, y = 2^(-x), etc. That may give you ideas about the base and the exponent,

- anonymous

oh well I was taught to use the first point first to find out what a equals for the b equation.

- anonymous

....? plotting?

- anonymous

Anytime you have x = 0 for an exponential equation, the y value for that point is the value of a.

- anonymous

Because any base raised to the 0 power equals 1.

- anonymous

still no comprendo senor.....or senora. You don't use the second equation for a.......gawh I wish math was simple for fooligans like me.

- anonymous

Ok. Lets break it down. We have 2 points (-2,9) and (0,1)
That gives 2 equations:
\[9 = ab^{-2}\]
and
\[1 = ab^0\]

- anonymous

With me so far?

- anonymous

si.

- anonymous

Ok. So what is
\[b^0=?\]

- anonymous

this is where your throwing me off. Don't i have to use the first equation for a first?

- anonymous

It doesn't matter. You use whichever is easier. In this case the second equation is easier. This will become apparent when you understand what b^0 is.

- anonymous

okay. I just want to make sure. I have a test tomorrow and if i use your tricks I don't want to get penalized.

- anonymous

Nope. Not a trick.

- anonymous

okay well to me its a different way so continue senor.

- anonymous

If you really want we can do it your way.

- anonymous

no I'm open to your idea. your the smarter one here so continue.

- anonymous

Lets do it your way and then we'll do mine and see that we get the same thing.

- anonymous

okay sounds good to me. :0]

- anonymous

Are you familiar with this notation?
\[\rightarrow\]

- anonymous

does that mean equals or references to the next step?

- anonymous

not equals. It means 'implies', 'therefore', or 'leads to'

- anonymous

So something like
\[ A = 5\]
\[ A +B = 10 \rightarrow 5 + B = 10 \rightarrow B = 5\]

- anonymous

Make sense?

- anonymous

okay continue :0]

- anonymous

So using the first equation and solving for a:
\[9 = ab^{-2}\]
\[\rightarrow 9b^2 = ab^{-2}b^2\]
\[\rightarrow9b^2 = ab^0 \rightarrow a = 9b^2\]

- anonymous

Follow that?

- anonymous

Okay question time! Don't you divide by b^2? and keep 9/b^2 for the next equations a? for example:
1=(9/b^2)(b^0)

- anonymous

then solve for b and then plug in b for the original equation.

- anonymous

Ah. No because
\[b^{-2} = \frac{1}{b^2}\]

- anonymous

So you need to multiply by b^2 in order to get a by itself, not divide.

- anonymous

oooooh so thats where I missed the mark. Okay i get it so if i have a negative, instead of dividing I multiply.

- anonymous

Think about it this way. You want to move the b over to the other side. You know that multiplying powers of the same base you add exponents right?
\[a^b * a^c = a^{(b+c)}\]

- anonymous

yes and when you divide you subtract.

- anonymous

Right. So here's the million dollar question. What is
\[k^0\]

- anonymous

0 right?

- anonymous

Nope.
2^4 = 16
2^3 = 8
2^2 = 4
2^1 = ?
2^0 = ?

- anonymous

shoot I change my mind its 1!! now wheres my money?

- anonymous

heheh.

- anonymous

Right. So if we have
\[ab^2 = 9\]
We can multiply both sides by b^-2
\[a*b^2*b^{-2} = 9*b^{-2}\]

- anonymous

And we end up with
\[a*b^{2+ (-2)} = 9b^{-2} \]
Which simplifies to
\[a*b^{0} = 9b^{-2} \]
Which simplifies to
\[a*1 = 9b^{-2} \]
Which simplifies to
\[a = 9b^{-2} \]

- anonymous

Now that wasn't the equation we had. We had
\[ab^{-2} = 9\]
So instead of multiplying by b^-2 we need to multiply by b^2

- anonymous

okay gotcha.

- anonymous

So we end up with
\[a = 9b^2\]

- anonymous

And we plug that into the second equation.

- anonymous

So what do you get when you plug in 9b^2 for a in the second equation.

- anonymous

i got 1/9=b^2

- anonymous

after dividing 9 from 9b^2 of course.

- anonymous

So b is?

- anonymous

b=1/3 Like I had before...

- anonymous

Yeah. So now plug that in for b in either of the two equations.

- anonymous

then i got a=1. and a full answer of y=1(1/3)^x

- anonymous

Yep.

- anonymous

So now look at what happens when you plug in 1/3 for b in the second equation. Does it matter what b was?

- anonymous

what do you mean? I got the whole equation right didnt i?

- anonymous

Yes. I'm trying to show my original idea.

- anonymous

oh okay. well no it doesn't matter what b was. So yes you were right. hahaa that all you wanted to hear wasnt it?

- anonymous

\[1=ab^0\] is our second equation from the beginning.
You plug in 1/3 and you get a = 1. But does it matter what you plug in for b?

- anonymous

no just as long as the answer is correct. :0]

- anonymous

Right. So you can see that for exponential equations if you are given a point (0,k). You already know that a must be k.

- anonymous

oh okay yeah I get it.

- anonymous

Because
\[ab^0= k\rightarrow a*1 = k \rightarrow a = k\]

- anonymous

Which means you can jump right to finding b and save some time.

- anonymous

so if i do it either way i will end up with the right answer. Okay i'm getting your drift here. thanks a bunch! your so patient and I feel more confident it passing my test. SO thanks it really means a lot!!

- anonymous

Anytime =)

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