Question

Determine wether the function represents exponential growth or decay. Then find the y-intercept.
1. y=4(5/6)^x
a. exponential growth;4
b. exponential decay; 4 (my choice)
c. exponential growth; 0
d. exponential decay; 5

2. y=12(17/10)^x
a. exponential growth; 1.7 (my choice)
b. exponential decay; 1.7
c. exponential growth; 12
d. exponential decay; 12

3. y=129(1.63)^x
a. exponential growth; 1.63
b. exponential decay; 1.63
c. exponential growth; 129
d. exponential decay; 129

Answer 1

For number 3 my choice was A

Answer 2

Exponential function is in the form: \(\large\color{#000000 }{ \displaystyle y(x)=a(b)^x }\)

Note that, when \(\large\color{#000000 }{ \displaystyle x=0; }\) \(\large\color{#000000 }{ \displaystyle y(0)=a(b)^0=a\cdot 1=a }\)
So, \(\large\color{#000000 }{ (0,a) }\) is the y-intercept of this exponential function.

Also, note that if \(\large\color{#000000 }{ \displaystyle b=1; }\) \(\large\color{#000000 }{ \displaystyle y(x)=a(1)^x=a\cdot 1=a}\)
THEN, this is simply a horizontal line \(\large\color{#000000 }{ \displaystyle y=a }\)

And, note that if \(\large\color{#000000 }{ \displaystyle b=0; }\) \(\large\color{#000000 }{ \displaystyle y(x)=a(0)^x=a\cdot 0=0}\)
THEN, this is simply a horizontal line \(\large\color{#000000 }{ \displaystyle y=0 }\) (or "x-axis).
\(\Huge \color{#000000 }{ _\text{__________________________} }\)

Answer 3

And now let's see what happens if \(\large\color{#000000 }{ \displaystyle 0<b<1 }\), and if \(\large\color{#000000 }{ \displaystyle b>1 }\).

Answer 4

Okay i've got it so far

Answer 5

let's for instance choose b=2, and for now will stick to positive a (say a=3).
\(\large\color{#000000 }{ \displaystyle y(x)=3(2)^x }\)

What happens as x grows?
\(\large\color{#000000 }{ \displaystyle x=1 \quad \quad y(1)=3(2)^1=3(2)=6 }\)
\(\large\color{#000000 }{ \displaystyle x=2 \quad \quad y(2)=3(2)^2=3(4)=12 }\)
\(\large\color{#000000 }{ \displaystyle x=3 \quad \quad y(3)=3(2)^3=3(8)=24 }\)
\(\large\color{#000000 }{ \displaystyle x=4 \quad \quad y(4)=3(2)^4=3(16)=48 }\)

Answer 6

So the function keeps growing × 2 ...
And you can try after this problem and
choose any b and a that satisfy,
a>0
b>1

and you will get that the function will grow on and on with no limit.

Answer 7

So, in fact, any time
a>0
b>1
then \(y(x)=a(b)^x\)
is EXPONENTIAL GROWTH,
with y-intercept at y=a

Answer 8

Now, if 0<b<1,

Let's choose
b=1/2
a=4 (let's still stick to positive a)

\(\large\color{#000000 }{ \displaystyle y(x)=4\left(\frac{1}{2}\right)^x }\)

\(\large\color{#000000 }{ \displaystyle x=1 \quad \quad y(1)=4\left(\frac{1}{2}\right)^1=4\left(\frac{1}{2}\right)=2 }\)

\(\large\color{#000000 }{ \displaystyle x=2 \quad \quad y(2)=4\left(\frac{1}{2}\right)^2=4\left(\frac{1}{4}\right)=1 }\)

\(\large\color{#000000 }{ \displaystyle x=3 \quad \quad y(3)=4\left(\frac{1}{2}\right)^3=4\left(\frac{1}{8}\right)=1/2 }\)

\(\large\color{#000000 }{ \displaystyle x=4 \quad \quad y(4)=4\left(\frac{1}{2}\right)^4=4\left(\frac{1}{16}\right)=1/4 }\)

Answer 9

So the function keeps decaying and it will approach zero closer and closer the bigger x-value we plug into the function.

Answer 10

So, you can probably tell that
when \(0<b<1\), (and when a>0)
the function is going to decay...
(called "exponential decay")

and as I showed in the beginning, the y-intercept is (still) y=a.

Answer 11

Bottom line:
\(\large\color{#000000 }{ \displaystyle y(x)=a(b)^x }\)

y-intercept: y=a
_______________________
■ If b>1, the function is "exponential growth".
■ If 0<b<1, the function is "exponential decay"

Answer 12

your y-intercepts in #2 AND #3 are incorrect.

Answer 13

They are both C if I am thinking correctly?

Answer 14

@SolomonZelman That was really helpful thank you

Answer 15

yes, #2 and #3 are both C, and #1 --> B was correct....

Answer 16

So,

B, C, C; respectively.

Answer 17

good luck!

Answer 18

Thank you so much! @SolomonZelman