Question

4=a^-2
What is the next step?

Original question: Give a possible formula for a function (-1,8) (1,2)

Answer 1

5-3.x

Answer 2

try linear first

Answer 3

8=Qnaut a^(-1)
over top of
2=Qnaut a^(1)

Qnaut cancels top and bottom, and a^1 is brought up top and its a^-1(-1) on top now.
8/2 is 4 so I got
4=a^-2

Answer 4

Its in the family of exponential functions

Answer 5

It maybe a quotient of some sort

Answer 6

This is review in the beginning of a Calc class

Answer 7

The answer is y=4(1-2^-x)

Answer 8

I guess so ? I'm really trying hard to get better at math
So Im not sure about linear or not.

Answer 9

ok let me see

Answer 10

I have just been working with point slope form too, but I can't do it!

Answer 11

oh crap hold on I forgot I could get M that way

Answer 12

Im getting y=-3x+6

Answer 13

8-2/-1-1
=-3
which is M correct?

Answer 14

y=-3x+5 :)
ok now how does this eventually equal
y=4(2^-x) the answer given in the book

Answer 15

what I described up top with Qnaut
thats how you solve these I did some in my Math Lab at school already

Answer 16

the question is give a possible formula for those two points

Answer 17

just a variable like x

Answer 18

ok thats where I was

Answer 19

still doesnt match book answer though

Answer 20

where did that go, I was trying 1/a^2
that seems like a good idea

Answer 21

ok yes I can get a = +/- 1/2 now

Answer 22

isn't the function

\[f(x) = \frac{1}{2^{(x - 2)}}\]

that satisfies the the 2 points that you have

Answer 23

the answer in the book is
y=4(2^-x)

Answer 24

oh and my solution

\[4a^2 = 1\]

so

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

\[a = \pm \frac{1}{2}\]

Answer 25

well look at 4 its 2^2

so you have

\[y = 2^2(2^{-x}) \]

then

\[y = 2^{2 - x}\]

Answer 26

where does the 1/2 come in from solving for a?

Answer 27

lol...what is the square root of 1/4 ...?

Answer 28

Since the exponential formula
y = Po a^x
From (-1,8) (1,2) :
y2/y1 = a^x2/ a^x1
= a ^ ( x2 - x1)
8/2 = a^(-2)
4 = 1/a²
=> a = ± 1/2

2 = Po ( 1/2)^1
=> Po = 4
Now plug into:
y = Po a^x
= 4 (1/2)^x

Answer 29

@campbell_st It takes me forever to understand what the poster need :'(

Answer 30

how do you get from solving for a
to y=2^2 equation?

Answer 31

how do I work the 1/2 into the final equation?

Answer 32

well you said the textbook answer is

\[y= 4(2^{-x})\]

and

\[4 = 2^2\]

then using the index laws for multiplication.... add the powers it can be written as

\[y = 2^{2 - x}\]

Answer 33

you don't need to work 1/2 into your answer as

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

Answer 34

ok got it!
man this is tough for me to see the step I need more practice

Answer 35

and thats why

\[f(x) = \frac{1}{2^{x - 2}} \]

fits your points

Answer 36

yep... lots of work on index laws....

Answer 37

index laws ok now I know what they are called at least
thanks guys!

Answer 38

Actually, it's not difficult as it looks
The first step is using ratio => a
The second step is plug a and 1 point into the exponential formula => Po
That's all :)

Answer 39

You make it difficult because you didn't point out find the exponential formula!