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Transforming Linear Functions

Let g(x) be the indicated combined transformation of f(x)=x. Write the rule for g.

horizontal stretch by a factor of 5 followed by a horizontal shift right 2 units.



That’s what the problem says. I don’t really understand how you transform linear functions at all. I don’t know if I said enough in the above question for you to help me with it or not ^^;

horizontal stretch of a graph by a factor of n makes f(x) as f(x/n)

since your graph is stretched by a factor of 5, your f(x) is transformed to f(x/5) = x/5

a horizontal shift of n units right transforms f(x) to f(x-n)

in your graph, it is 2 units, so your function becomes f(x-2) = (x-2)/5

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Let g(x) be the indicated combined transformation of f(x)=x. Write the rule for g.

horizontal stretch by a factor of 5 followed by a horizontal shift right 2 units.

That’s what the problem says. I don’t really understand how you transform linear functions at all. I don’t know if I said enough in the above question for you to help me with it or not ^^;

horizontal stretch of a graph by a factor of n makes f(x) as f(x/n)

since your graph is stretched by a factor of 5, your f(x) is transformed to f(x/5) = x/5

a horizontal shift of n units right transforms f(x) to f(x-n)

in your graph, it is 2 units, so your function becomes f(x-2) = (x-2)/5

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Can you help with word problems?

1.) Train A goes 12mph slower than Train B. Train A travels 230 miles in the same time that Train B travels 290 miles. Find the speed of each train.

2.) Zeno takes 3 hr longer to paint a floor than it takes Lia. When they work together, it takes them 2 hrs. How long would each take to do the job alone?

3.) A boat travels 15 km/h in still water. If the boat travels 140 km downstream in the same time it takes to travel 35 km upstream. What is the speed of the river?

I’m not sure what sort of equations to use in order to set up the problem..

Any help would be great, thanks :)

very simple

1) let speed of train A is Va, B be Vb

given, Va = Vb - 12………………eqn 1

speed= distance/time

time taken is same for each of the train, let it be t

given, distance traveled by A/ speed of A = distance traveled by B/speed of B…just equating time

or

230/Va = 290/Vb

put Vb= Va + 12 from equation 1 and solve 

2) Let time taken by Lia to paint a floor be x hours.

Thus, time taken by Zeno would be x + 3 hours.

This simply means when alone, in 1 hour,

Lia completes 1/x of the floor, while

Zeno completes 1/(x+3) of the floor.

If working together, in 1 hour,

Both will complete 1/x + 1/(x+3) of the floor

Since if working together, it takes them 2 hours to complete the job, in 1 hour they would complete 1/2 of the whole floor, or in other word:

1/x + 1/(x+3) = 1/2

[(x+3) + x] / [(x)(x+3)] = 1/2

(2x+3)/(x^2+3x) = 1/2

2(2x+3)=(x^2+3x)

4x+6=x^2+3x

x^2-x-6=0

Solving for x, x=3 hours or x=-2 (Not acceptable, as time taken to paint a floor can’t be negative)

3) Let F = the downstream speed of the water.
Then the boat’s upstream speed is: 15 - F
The boat’s downstream speed is: 15 + F
Assume both the journeys mentioned take T hours, then using “speed x time = distance” we get:
Downstream journey: (15 + F)T = 140
Upstream journey: (15 - F)T = 35
Add the two formulae together:
(15 + F)T + (15 - F)T = 140 + 35
15T + FT + 15T - FT = 175
30T = 175
T = 35/6
Use one of the equations to find F:
(15 + F)T = 140
15 + F = 140/T
F = 140/T - 15
F = 140/(35/6) - 15
F = 24 - 15
F = 9
i.e. the downstream speed of the water is 9 kph
Therefore, the boat’s speed downstream is 15 + F = 15 + 9 = 24 kph.

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zeldahansonpotter: What am I doing wrong? Apparently you’re suppose to get xcosx+2sinx that is what you call a silly mistake and getting the answer wrong even after being thorough with concepts while calculating y”, you left out the cos x ….. you did not differentiated it. do it, you shall have your answer

zeldahansonpotter:

What am I doing wrong?
Apparently you’re suppose to get xcosx+2sinx

that is what you call a silly mistake and getting the answer wrong even after being thorough with concepts

while calculating y”, you left out the cos x ….. you did not differentiated it.

do it, you shall have your answer

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So will you math ed guys do me a favor?

kat523719:

Actually, any of you who would be willing please do it. I had to make this learning module for a class. Would you guys look at it and tell me what you think?

http://katsmathy.wordpress.com/learning-module/

it is very good, you used google docs to create your quizzes,

if you want to include grading too in google docs quizzes, like after a student finishes the test, he should be able to see his grades too, let me know.

or if you want to include multimedia too in the questions, let me know.

(Source: bug523719)

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okay two more questions and I think that’s it.

here i’m solving the logarithmic equations and eliminating any extraneous solutions:

log(x+5)-log(4x^2+5)=0

and then

this one i just need my answer checked on:

1/2[log(x^2-9)-log(x-3)]-logx

my answer came out to log (x+3)^1/2 over x.

the first one is quite simple

log(x+5) - log(4x^2 + 5) = 0

or

log[(x+5)/4x^2 + 5)] = 0

or (x+5)/(4x^2 + 5) = 1………………since log 1 = 0

or x+5 = 4x^2 + 5

0r 4x^2 - x = 0

or x(4x-1)= 0

or x = 0 or x = 1/4

second one, in some time, typing from phone and its getting really hectic to type with powers. 

be back in some time

thank you so much by the way! this is awesome and now I am finally actually understanding my math, which is good. (no more failing tests for me [hopefully]!)

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actually i have a couple i need help with.

the problem is 10^x=2^(-x+4)   I have to solve it using logs.  I can get it to the point  xln5=(-x+4)ln2  but i’m not really sure where to go from there.

10^x = 2^(-x+4)

take log of both sides, to base 10

xlog10 = (-x + 4) log 2

x[ log (2*5)] = (-x + 4) log2

x( log 2 + log 5) = -xlog2 + 4log 2

x log 2 + xlog5 = -xlog2 + 4log2

x (log 2 + log 5 + log2) = 4log2

x = 4log2/(log 2 + log 5 + log 2)

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Hi can you find the derivative of this function and simplify the answer? y=(2x^3 +9x)^5 

( please tag ‘chikaras’ so I’ll be sure to see the answer :) )

y = (2x^3 + 9x)^5

let 2x^3 + 9x = u ………..y = u^5, dy/du = 5u^4

so (6x^2 + 9)= du/dx

dy/dx = dy/du * du/dx……………using chain rule

dy/dx = 5u^4 * (6x^2 + 9)

substitute value of u = 2x^3 + 9x

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A Girl Unfathomed...: Math rant.

theunfathomed:

I study hours and hours on math. I go to tutoring, I do the labs, I find time for more notes, but NOTHING IS WORKING AND IT’S DRIVING ME INSANE! I’m trying so hard to pass this course! WHY DO PEOPLE WHO HARDLY PUT IN THE EFFORT JUST SLIDE ON BY WHILE PEOPLE WHO PUT WORK INTO EVERYTHING HAVE TO…

I have some spare time right now, if you can tell what exactly you find difficult, I can help you.

I don’t like some one who is doing so much efforts on learning math to feel bad about it.

i will help you if you want

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