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# Reciprocal graphs watch

1. I am so confused that I do not know how to even start

Can someone explain how to do this, please?

2. (Original post by TheRealLifeBane)
I am so confused that I do not know how to even start

Can someone explain how to do this, please?

horizontal asymptotes
think what happens to y as x gets very large and positive (or negative)

vertical asymptotes
they occur when division by zero takes place
3. Thanks.

I searched online for how to obtain the horizontal and vertical asymptotes. However, I have never learnt this so I might be wrong.

Therefore, I am assuming for part (a) I have to find the asymptotes of y=f(x) and then just find the transformation right?

If so, I did this to get the VA:
x-1=0 Hence x=1, which is why x will never equal 1
So y=f(x-3) means x=1 becomes x=1+3 which is x=4

And for the HA:
Using 1x^1/(1x^1-1), the highest powers are equal so I use the leading coefficients 1/1=1
Then, since y=f(x-3) will not affect the y values as it is in the brackets, y=1 will remain as y=1

In part (b) I made y=f(x) equal to y=f(x-3)So x/(x-1)=(x+3)/(x-1+3)
Eventually, I got 0=-3 but since that is impossible I thought no solutions meant there are no points of intersections.

Now, did I get this question right, partially right or was I completely wrong?
4. (Original post by TheRealLifeBane)
Thanks.

I searched online for how to obtain the horizontal and vertical asymptotes. However, I have never learnt this so I might be wrong.

Therefore, I am assuming for part (a) I have to find the asymptotes of y=f(x) and then just find the transformation right?

If so, I did this to get the VA:
x-1=0 Hence x=1, which is why x will never equal 1
So y=f(x-3) means x=1 becomes x=1+3 which is x=4

And for the HA:
Using 1x^1/(1x^1-1), the highest powers are equal so I use the leading coefficients 1/1=1
Then, since y=f(x-3) will not affect the y values as it is in the brackets, y=1 will remain as y=1

A cursory glance says that's all fine.

In part (b) I made y=f(x) equal to y=f(x-3)So x/(x-1)=(x+3)/(x-1+3)
Eventually, I got 0=-3 but since that is impossible I thought no solutions meant there are no points of intersections.

Which looks like a quadratic begging to be solved.

Now, did I get this question right, partially right or was I completely wrong?
5. May be worth writing the function as
(x-1)/(x-1) + 1/(x-1)=1+1/(x-1).
It may make it easier to see what's going on.

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Updated: February 15, 2016
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