Car depreciation problem

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  • #1
jai6638
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URGENT Basic Calc Help needed

Q3) If a car depreciates continuously at a rate of 23.5%, when ( years ) will a 25,000 dollar car be worth $12,000 ?

A= pe^rt
(12,000)/(25000)= e^(.235t)
Ln ( 12,000/25000)=.235t
T= .0038

This answer does not seem right. Where am I going wrong?

Q2) Find the vertical and horizontal Asymptotes of f(x) = (3x) / ( 2x-4)

Vertical Asymptote = 2
Horizontal Asymptote= dunno.. I can't use limits here directly since the exponents of X are the same. If i tried dividing them so as to get a proper form, it doesn't seem to work out.



Q3)

Find the equation of the normal line at x=2 of the function f(x) = x^2-5x+4

X=2
Y=-2 ( plugged in 2 into the function )

Y=mx+b
-2=2m+b

How would I find the slope and B? Is the slope -1?


Q4) Find the average rate of change between 2 and -1 of

f(x) = x^2+ 3x

I found the derivative to be 2x+3 and plugged in 2 and -1 into the derivative to get -2 and 10.

10=11/2 +b = 9/2 = b

Is this the answer? y= -11/4x + 9/2
 
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Answers and Replies

  • #2
TD
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1) I think this should work: [tex]25000\left( {1 - 0.235} \right)^x = 12000[/tex]

2) Your VA is correct but why can't you use limits for the HA? If the exponent in the nominator is one more than the one of the denominator, you may have any line as asymptote but HA appear when the highest exponents are equal.
Try taking the limit to + and - infinity.

3) If you can find the tangent, you're almost finished, since you just have to take the perpendicular line then. (with m' = -1/m)

4) Don't you need to find a numerical answer? I'd say the avarage change between x = a and x = b would be (f(b)-f(a))/(b-a)
 
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  • #3
jai6638
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2) Your VA is correct but why can't you use limits for the HA? If the exponent in the nominator is one more than the one of the denominator, you may have any line as asymptote but HA appear when the highest exponents are equal.
Try taking the limit to + and - infinity.

2) well the textbook said that I can't use limits for HA when the exponent of X in num and den are equal. Anyhow, I tried to do so, and the X cancel out thus giving me a limit of 3/2 .. is that correct?

1) It is bein compounded continuously so how can I use the standard thre equation that you listed. However, I tried it anyways and got an answer of 2.73 years... don't know if this is correct.
 
  • #4
jai6638
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For Question 3: is y=2x-5 the answer? ( i plugged in 2 into the derivative of the equation and then put it in point soope form )
 
  • #5
TD
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jai6638 said:
2) well the textbook said that I can't use limits for HA when the exponent of X in num and den are equal. Anyhow, I tried to do so, and the X cancel out thus giving me a limit of 3/2 .. is that correct?

1) It is bein compounded continuously so how can I use the standard thre equation that you listed. However, I tried it anyways and got an answer of 2.73 years... don't know if this is correct.

For Question 3: is y=2x-5 the answer? ( i plugged in 2 into the derivative of the equation and then put it in point soope form )
2) Yes, you have a HA (both at the right as left) at y = 3/2

1) I wasn't 100% sure about that one, someone else may verify.

3) That doesn't seem correct. Your slope at (2,f(2)) is -1 so the perpendicular direction is 1. So the line should be y - f(2) = 1*(x-2)
 
  • #6
jai6638
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y - f(2) = 1*(x-2)

that cnat be the final equation ?

Is the final equation y= 2x since Y seems to equal 2 according to the equoted equation and if you plug it in point slope form ( 2-2= b ) you would get y=2x.
 
  • #7
TD
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y = f(2) would be -2. So the line becomes: y = x - 4
 
  • #8
jai6638
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hmm I see.. so you don't have to find a derivitive eh?

thanks .. need some help with the other answers though
 
  • #9
TD
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I did use the derivative to find the slope (of the tangent line).
If that slope is m, the perpendicular slope of the normal line is m' = -1/m.
 
  • #10
jai6638
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wont the perp slope be (-1/3) then?

Damn I am confused. I'd really appreciate it ifyou could please write down the steps for what you did for q3..
 
  • #11
The Bob
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jai6638 said:
Damn I am confused. I'd really appreciate it ifyou could please write down the steps for what you did for q3..
Would you mind if I tried?

The Bob (2004 ©)
 
  • #12
jai6638
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sure....
 
  • #13
The Bob
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[tex]f(x) = x^2 - 5x + 4[/tex]
[tex]f(2) = 2^2 - (5 \times 10) + 4 = 4 - 10 + 4 = -2[/tex]
We now have the place where the normal line will cross, at (2, -2).

Differentiate the function to get:

[tex]f'(x) = 2x - 5[/tex]

Put in the value of x, which is 2, and we get the gradient at that point on the curve:

[tex]f'(2) = (2 \times 2) - 5 = -1[/tex]

To find the perdendicular gradient to this, we have to find the missing value (as we know that f'(2) is equal to m1):

[tex]m_1 \times m_2 = -1[/tex]

[tex]-1 \times m_2 = -1[/tex]

[tex]m_2 = \frac{-1}{-1} = 1[/tex]

We now have everything we need for the second equation of a straight line:

[tex]y - y_1 = m (x - x_1)[/tex]

We know that y1 is -2 and that x1 is 2. We now know that m (the gradient) is 1 (for the normal line to that equation at x = 2).

Plug it all in and you get:

[tex]y - y_1 = m (x - x_1)[/tex]

[tex]y - (-2) = 1 (x - 2)[/tex]

[tex]y + 2 = x - 2[/tex]

[tex]\Rightarrow y = x - 4[/tex]

I hope that all makes sense. Do say if not.

The Bob (2004 ©)
 
  • #14
jai6638
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I'm relatively new to differenciation so i'd like to know why are we finding the perpendicular gradient since we want to find the value at a particular point?

Also, how is (m1)(m2)=-1?

THanks much. really appreciate it guys..
 
  • #15
The Bob
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jai6638 said:
I'm relatively new to differenciation so i'd like to know why are we finding the perpendicular gradient since we want to find the value at a particular point?

Also, how is (m1)(m2)=-1?

THanks much. really appreciate it guys..
Sorry for the late reply. :blushing:

The differentiated line is the tangent to the line. The normal line, which you want, is perpendicular to the tangent of that point.

The way to work out the perdendicular gradient is (m1)(m2) = -1. Take the line y = x. The gradient is going to be 1. The line perpendicular to this one is y = -x (through the origin); but why? That equation. To the m's multiply to -1, the other m value (as one of them is 1) must be -1. The same is for your posted question. It is then used in the Second Equation for Straight Lines.

Did any of that make any sense at all?

The Bob (2004 ©)
 
  • #16
jai6638
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Yeah I did... essentially, its coz the slope of a perp line is the negative inverse ( -1/m ) eh?

Cool.. thanks much.. Could you also confirm the other answers please?
 
  • #17
The Bob
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jai6638 said:
Yeah I did... essentially, its coz the slope of a perp line is the negative inverse ( -1/m ) eh?

Cool.. thanks much.. Could you also confirm the other answers please?
Ok, but bear in mind that I am very tired and that I am going to bed after this post.

1. TD knocked that one on the head in post 2
2. [tex]x = 2[/tex] and [tex]y = \frac{3}{2}[/tex]
3. y = x - 4
4. I believe it to be 4 but that might be wrong.

Now I must sleep. It is 13 minutes to 1 in the morning here. :smile:

The Bob (2004 ©)

P.S. Hope that helps.
 
  • #18
jai6638
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thanks much

Can anyone confirm the answer for question 4?
 
  • #19
TD
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jai6638 said:
thanks much

Can anyone confirm the answer for question 4?
Using (f(b)-(a))/(b-a) as I said, I agree with Bob that is should be 4.

I had to log off last night but I see Bob helped you further, hope it's all clear now :smile:
 
  • #20
jai6638
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I tried puttig it in that form and i got -2+10 / 2+1 = 8/3

doh.. need to go for my exam now... thanks anyways.
 
  • #21
TD
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jai6638 said:
I tried puttig it in that form and i got -2+10 / 2+1 = 8/3

doh.. need to go for my exam now... thanks anyways.
f(x) = x^2+ 3x between 2 and -1

f(-1) = -2
f(2) = 10

=> (f(2)-f(-1))/(2-(-1)) = (12)/(3) = 4

Good luck with your exam!
 
  • #22
jai6638
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Thanks much... damn I am trying to call my school office and no one picking up.. ! I hope they arent taking a day off or something!
 
  • #23
TD
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It would give you more time to prepare, not your fault ;)

If you have any more questions, don't hesitate to ask for help!
 
  • #24
jai6638
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finally got through... going to take it at 1 pm... thank god!


thanks.
 
  • #25
TD
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Good luck then!
 
  • #26
jai6638
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Write the function in linear factorization form:

f(x)= 2x^5-3x^4 - 15x^3 + 12x^2 + 22x - 12

How do i do this? I graphed it but couldn't find any real zeros to start off with...

EDIT: nevermind.. -2 is a zero. however, once I perform syntehtic division and get another equatoin with the highest degree being 4, how do i know what zero
to use next since -2 is the only zero listed on the calc?
 
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  • #27
TD
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Possible zero's are the divisors of the constant, here 12.
So try: 1,2,3,4,6,12 and their negatives.

You can already cancel out 1 and -1 thanks to two simple rules:
- divisibility by (x-1) <=> sum of all coefficients is 0
- divisibility by (x+1) <=> sum of all coefficients of the even powers = sum of all coefficients of the odd powers.

Intuitively, you cancel out the larger ones too (4,6,12) since the power 5 is too large for those to be zeros. Try 2,-2,3,-3 :wink:
 
  • #28
TD
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Didn't see your edit. I see you found the zero -2, so you can divide by (x+2). You should get:

[tex]2x^5 - 3x^4 - 15x^3 + 12x^2 + 22x - 12 = \left( {x + 2} \right)\left( {2x^4 - 7x^3 - x^2 + 14x - 6} \right)[/tex]

Now try the same on the new 4th degree polynomial, the divisors of the constant (here 6).
For the same reason as above, you cancel 1 and -1, leaving 2,-2,3,-3 as possibilities.
 
  • #29
jai6638
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So you use the rational root theorm eh? I forgot about that.. Thanks..
 
  • #30
TD
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It's not certain that you'll find roots like that, but IF there are integer roots, then they'll be divisors of the constant. When no divisor of the constant is a zero, there still may be radical or complex zeros.
 
  • #31
jai6638
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How would you find the complex zeros or radical in this case ( if there were not integer roots ) ? would i have more information if i had to find only complex zeroes?

Also, there was a questoin in a sample test where it said " Explain how you would find non-real roots in an equation using your calculuator".. Would I use the table and find it since the graph won't display complex roots?
 
  • #32
TD
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I'm no specialist with (graphical) calculators but finding roots (whether real or complex) is always possible using formula until the degree of 4.

Well-known is the one for quadratic equations, using [tex]\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}[/tex] where you get complex solutions if [itex]{b^2 - 4ac}[/itex] is negative.
 
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  • #33
jai6638
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Find two triangles for which A=36 degrees , a= 16 and C=17

So do I first find all the angles and sides for the first triangle? What do I do next? Divide the triangle into two and then find the 2 angles of the divided triangle?
 
  • #34
TD
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Did you mean C = 17° the angle or c = 17 the side?
 
  • #35
jai6638
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i think c is the side.
 

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