Question

ok i have two problems but both have the same question. help please write the equation for the translation of y=4/x that has the given asymptotes. 1) x=0, y=3 2) x=-3, y= -4

Answer 1

help @whpalmer4

Answer 2

@jdoe0001

Answer 3

We want to translate \(y = 4/x\) so that it has asymptotes \(x=0, y=3\).

Answer 4

What are the asymptotes of \(y = 4/x\) before translation?

Answer 5

its ok. i now this is smiler to the other problems but. i dont know how to work backwards.

Answer 6

is it x=0 and y=0 ????
im still a little rusty on this asymptote stuff :)

Answer 7

Yes. The fact that we have a 4 in the numerator instead of 1 doesn't change the asymptotes, because \(x\) is going to huge numbers (or very small numbers) and it washes out the effect of the 4.

So if the one asymptote remains \(x=0\), are we shifting the graph up or down, or right or left?

Answer 8

y = 3 is a horizontal or vertical line?

Answer 9

sorry i said that wrong lol ummm what i was trying to say was is the graph going to move up to the 3 on the x axis?

Answer 10

Yes, the graph will move up by 3 units. What do we have to do to \(y = f(x)\) to make it move up by 3 units?

Answer 11

(doesn't matter what \(f(x)\) is, the process is the same)

Answer 12

we will add 3

Answer 13

right??

Answer 14

Indeed we will!

So what is our new function?

Answer 15

is it y=4/x +3???

Answer 16

Is that \[y = \frac{4}{x}+3\] or \[y = \frac{4+3}{x}\]?

Answer 17

the first one.............is that right???? i hope so :/

Answer 18

it will be the first one i graphed it to make sure. right??

Answer 19

@whpalmer4

Answer 20

yes, the first one is correct.

Sorry, my sweetie is watching "Groundhog Day" and we just got to a particularly funny scene :-)

Answer 21

oh ok its not a problem i understand.

Answer 22

ok so for part 2) what will that one be. its harder than this first one

Answer 23

it is x=-3 and y= -4

Answer 24

i know we will subtract 4 from the equation.

i know it will look like this if x was 0.......... y=4/x -4

Answer 25

@whpalmer4

Answer 26

Okay, yes, subtracting 4 will shift the horizontal asymptote down to y = -4. Now we need to shift the graph sideways so the vertical asymptote is at x = -3.

remember \(y=1/x\)? Where was the vertical asymptote? What happened there when you tried to compute \(y\)?

Answer 27

aww man i dont remember.

Answer 28

dang it!!!!! i thought i was doing so well

Answer 29

The vertical asymptote was at \(x=0\), right? what happens when we plug \(x = 0\) into \(y = 1/x\)?

Answer 30

yes ok now i remember. we didnt get anything there was a error

Answer 31

Right. So we need to have an error like that at x = -3. How do we change our function to do that?

Answer 32

im not sure how to do that

Answer 33

what caused the error?

Answer 34

when we divided by 0

Answer 35

Okay, how do we divide by zero if x=-3?

Answer 36

um i know if we that -3 and 3 cancel each other out??? if that helps any.....

Answer 37

yes. it does.

we want to have a 0 in the denominator when x = -3.

\[x+a = 0\]\[x = -3\]\[-3+a = 0\]
any ideas?

Answer 38

a=3 right???

Answer 39

Yes. so what is our new denominator? remember, it will equal 0 when x = -3

Answer 40

umm -3+3

Answer 41

can you write an expression involving \(x\) that =0 when x = -3?

Answer 42

oh maybe 3x???

Answer 43

3(-3) = -9, not 0

Answer 44

just using addition or subtraction

Answer 45

you're going to want to hit yourself in the forehead when you see the answer :-)

Answer 46

dang i thought i had it then i tested it and saw i was wrong is it 0x????

Answer 47

omg it will be x+3

Answer 48

im a retard lol i did hit myself 0.0

Answer 49

\[y=\frac{4}{x+3}-4\]

What is the value of that at \(x = -3\)?

Answer 50

-_- so bummed at myself

Answer 51

"watch out forehead, here comes the palm!" :-)

Answer 52

lol yep i should have known that

Answer 53

so, to give the function a vertical asymptote at \(x = a\), just make sure there is a \((x-a)\) product term in the denominator

Answer 54

wait what???

Answer 55

we wanted one at \(x = -3\), so we needed \((x-(-3)) = (x+3)\) in the denominator.

Answer 56

ok

Answer 57

continue

Answer 58

as a more general discussion, if we have a function \(y = f(x)\):

\(y = f(x) + a\) shifts the graph up/down by \(a\) units
\(y = af(x)\) expands the scale of the graph by a factor of \(a\) (and inverts it if \(a < 0\))
\(y = f(x+a)\) shifts the graph left/right by \(a\) units
any or all of these can be combined to achieve multiple translations simultaneously

Answer 59

oh ok i get it..... wel somewhat

Answer 60

So we had \(y = 1/x\). Multiplying by \(a=4\) expanded the vertical scale by a factor of 4. Adding \(+3\) to the value of \(x\) (\(f(x+3))\) shifted the graph so that the vertical asymptote previously at \(x = 0\) is now at \(x = -3\).

Answer 61

One more case I didn't mention:
\(y = f(ax)\) expands/contracts the horizontal scale by a factor of \(a\)

Answer 62

omg your going to kill me lol...... i have been doing this all day!!!!! my brain is fried lol 0.0
:S

Answer 63

For the horizontal scale, if \(a < 1\) the graph spreads wider, if \(a > 1\) the graph is narrower. For vertical scale it works the other way. \(a>1\) makes the graph taller, \(a<1\) makes it shorter.

Yeah, it's a bit overwhelming, isn't it :-)

Answer 64

But save the page and come look at it again tomorrow and the next day.

Answer 65

ok so is the answer y=4/x+3 -4

just want to be done until tomorrow. ill be back dont worry. lol

=S

Answer 66

Close: if you are going to write it horizontally like that, you MUST use parentheses to indicate that the denominator is (x+3). \[y = 4/(x+3) - 4\]

Answer 67

ok i will thank you lol

Answer 68

will you be on here tomorrow???

Answer 69

Otherwise, the order of operator precedence says that we do multiplication and division before addition and subtraction, so \[y = 4/x+3 -4 \]turns into \[y = 4/x - 1 = -1 + 4/x\]and that isn't the same thing.

Answer 70

Oh, I probably will be, though I couldn't tell you exactly when.

Answer 71

ok well i will just keep a look out :D THANK YOU SO MUCH i couldnt have dont it without you =D

Answer 72

If you send me a message, I'll get an email to that effect and know to come looking for you.

Answer 73

ok i will do that. see you tomorrow