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

    Can someone explain how to do this, please?

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    (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?

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    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
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    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?
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    (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.
    Not sure about this one: \displaystyle \frac{x}{x-1} = \frac{x+3}{x+3 -1} \iff x(x+2) = (x+3)(x-1)

    Which looks like a quadratic begging to be solved.




    Now, did I get this question right, partially right or was I completely wrong?
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    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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