Pre-Calculus exponential functions help?

- anonymous

Pre-Calculus exponential functions help?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- anonymous

make an x/y table to get points and plot

- anonymous

|dw:1443201477463:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

where did the 1/4 and 1/2 come from?

- anonymous

\[2^{-1} = \frac{ 1 }{ 2^1 }=\frac{ 1 }{ 2 }\]\[2^{-2}=\frac{ 1 }{ 2^2 }=\frac{ 1 }{ 4 }\]

- anonymous

Oh okay

- anonymous

So what do I do with the points from the table then? Plot them?

- anonymous

yes

- anonymous

But that table was only for 2^x, right? I still need a table for 2^-x?

- anonymous

yes

- anonymous

do you know your transformations?

- anonymous

kind of

- anonymous

so what if g(x) = f(-x) what is g compared to f? in other words, g is what transformation of f?

- anonymous

reflection?

- anonymous

about what?

- anonymous

the y axis

- anonymous

very good!
so f(x) = 2^x and g(x) = 2^(-x) hint, hint, wink, wink...
if you know one and you understand transformations, you really know both

- anonymous

so after I graph 2^x, then I just reflect it about the y axis for the graph of 2^-x?

- anonymous

right?

- anonymous

So they're exponential functions because they increase or decrease exponentially?

- anonymous

yes but where is the variable x?

- anonymous

I don't understand what you mean

- anonymous

\(f\left(x\right) = y = 2^x\)

- anonymous

x is in the exponent

- anonymous

and that's what is changing

- anonymous

thus an "exponential" function

- anonymous

ohhh okay!

- anonymous

so what if we're given the graph and need to find the function? (I have an example)

- anonymous

http://www.purplemath.com/modules/expofcns.htm
exponential functions are also characterized by multiplying in a recursive relation...
\[f(x)=2^x=2\cdot 2^{x-1}=2\cdot f(x-1)\]
so if you have a graph, divide f(x)/f(x-1) to find the base

- anonymous

how do we know what f(x) is?

##### 1 Attachment

- anonymous

f(2) = 9, what is f(1)?

- anonymous

2?

- anonymous

that's what it looks like to me...
so f(2)/f(1) = base

- anonymous

so 9/2 is the base?

- anonymous

that's what i think

- anonymous

okay, then what's the exponent part?

- anonymous

whoa... wait

- anonymous

what?

- anonymous

f(1) = 2 <- this one is not correct
f(2) = 9
f(0) = 1
it's 2 tick marks but how much is each tick mark?

- anonymous

I thought it showed that each mark is 1...

- anonymous

if you ask me, f(1) = 3 look closely at the spacing of the vertical tick marks.
the first tick is about half of the subsequent ticks

- anonymous

there are 5 marks to 9... 1 + 2 + 2 + 2 + 2 = 9

- anonymous

besides (9/2)^1 = 9/2 \(\ne\) 2 and
(9/2)^2 = 81/4 \(\ne\) 9

Looking for something else?

Not the answer you are looking for? Search for more explanations.