Question

1) Sara's new car depreciates 8% each year.
1.Write an exponential function to model the depreciation each month.
2.Write an exponential function to model the depreciation each week.
3.Write an exponential function to model the depreciation each day.
4.What relationship is there between the amount of depreciation and the time interval measured?

2) Given that the domain is all real numbers, what is the limit of the range for the function f(x) = 42^x - 100?

3) The function f(x) = 200 x (1.098)^x represents a village's population while it is growing at the rate of 9.8% per year.
1.Create a table to show the village's population at 0, 2, 4, 6, 8, and 10 years from now.
2.Use your table to create a graph that represents the village's population growth.
3.When the population doubles from its current size, the village will need to dig a new water well. To the nearest half of a year, about how long before it is time for the village to dig the new well?

4) Jacob deposits $60 into an investment account with an interest rate of 4%, compounded annually. The equation, y = 60(1 + 0.04)^x, can be used to determine the number of years, x, it takes for Jacob's balance to reach a certain amount of money, y. Jacob graphs the relationship between time and money.
1.What is the y-intercept of Jacob's graph?
2.What is the equation of the asymptote?
3.If Jacob doesn't deposit any additional money into the account, how much money will he have in eight years? Round your answer to the nearest cent.

SolomonZelman answers all the question but the answer for the first question is at the end even thought we said it can be answered.

Answer 1

I have sat here for almost tow hours trying to figure out these problems but have gotten no where.

Answer 2

for #1 are you given the initial cost of the car?

Answer 3

no

Answer 4

then I can't really help with part 1, asides from telling you that your "common ratio" here (or the base of this exponential decay) is 0.92.

Answer 5

for part 2,
42^x is always a positive number, so it is at least a zero.
Now,
(42^x) -100 is (the same way) at least a -100, just like any
42^x+C is at least a C.

Answer 6

well, not that 42^x is ever a zero, but it approaches zero very closely when x is negative, so although it wouldn't be zero, you can say that it is all numbers greater than 0.

Same thing would apply when you have 42^x-100 , it would be all real numbers greater than -100.

Answer 7

\(\large\color{black}{ \displaystyle f(x) = 200 (1.098)^x }\)
for the table just plug in 2, 4, 6, and on... for x to find what the y-values are.

Answer 8

you can graph a couple of points and based on them make an approximate sketch, if you aren't allowed to use a graphing calculator.
(such as https://www.desmos.com/calculator )
(this is part 3, 2)

Answer 9

part 3, 3

\(\large\color{black}{ \displaystyle f(x) = 200 (1.098)^x }\)
200 is the initial population, for it to double it should be 400.
So, write
\(\large\color{black}{ \displaystyle 400= 200 (1.098)^x }\)
and find out at which x value (at which year) the population doubles.

Answer 10

(use logarithms for an exact answer, or you can guess it, or use http://www.wolframalpha.com/ --> a computing engine)

Answer 11

how do you feel so far?

Answer 12

I am understanding what you are saying so far.

Answer 13

yes? great! Many people on here do not.. that means you have a good background to the subject.

Answer 14

your equation for part 4 technically is:

\(\large\color{black}{ \displaystyle y=\left( 1+\frac{0.04}{1}\right)^{1\times x} }\)
this is a compound interest formula

but they simplified it down, good:)

Answer 15

left out the 60 , sorry. it is like this \(\large\color{black}{ \displaystyle y=60\left( 1+\frac{0.04}{1}\right)^{1\times x} }\)

Answer 16

\(\large\color{black}{ \displaystyle y=60\left( 1.04\right)^{x} }\)

Answer 17

this is after adding in parenthesis.

Now,
1) y-intercept?
set x=0 and sole for y.

2) asymptote?
usually an asymptote of b^x is x=0, because x is almost b^x approaches 0 as x approaches negative infinity. So the asymptote is y=0.

When it is 60(b)^x (for any positive base b) then asymptote is y=60

if you are calling an asymptote.

Answer 18

for part 4, 3
\(\large\color{black}{ \displaystyle y=60\left( 1.04\right)^{x} }\)
(x is number of years, and y is the outcome after a certain number of years)

amount after 8 years? Plug in 8 for x!
\(\large\color{black}{ \displaystyle y=60\left( 1.04\right)^{8} }\)
\(\large\color{black}{ \displaystyle y=? }\) (just a matter of a calculator)

Answer 19

but about #1, really contact your teacher, because it is wrong.

Answer 20

OK. Thank you so much. You helped me incredibly.

Answer 21

\(\large\color{black}{ \displaystyle y=a( b)^{x} }\)
is a normal exponential function.

when it relates to a cost, "a" is the previous cost,
and you are multiplying times b an additional time as x grows.

like when x (number of years) is 0, you are not multiplying a times b. The reason is because everything has just began (year 0) - nothing happened yet.
and when x is (for example) 1 (1 year since) then you multiply a times b once.

when x is 2, then you go a*b*b... and on.

Answer 22

without knowing an original cost "a" you can't write a function as they ask in part 1.

Answer 23

yw

Answer 24

I found the answer for the first question incase anyone needs it.

1.\[f(x)=(0.92^{\frac{ 1 }{ 12 }})^{12t}\] Which equals \[f(x)=0.993^{12t}\]
2. \[f(x)=(0.92^{\frac{ 1 }{ 52 }})^{52t}\] Which equals \[f(x)=0.998^{52t}\]
3.\[f(x)=(0.92^{\frac{ 1 }{365 }})^{365t}\] Which equals \[f(x)=0.999^{365t}\]
4.The shorter the time interval, the less visible the depreciation of Sara's car