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Math Help - Can anyone help me with these questions

  1. #16
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    Hi thanks for the answers and I got most of them right but I got a question.

    On Q3 what does the graph suppose to look like?

    And on Q7a how did you get +4 on 8x^-1 to get the value for c. because if x = 2 I got 1/8(2) so I got 1/16 because I thought the power of -1 meant 1 over.

    Thank You
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by novadragon849 View Post
    Hi thanks for the answers and I got most of them right but I got a question.

    On Q3 what does the graph suppose to look like?

    And on Q7a how did you get +4 on 8x^-1 to get the value for c. because if x = 2 I got 1/8(2) so I got 1/16 because I thought the power of -1 meant 1 over.

    Thank You
    i will check question 3 to see what you are talking about. as for your other question. you are right, -1 power means 1 over, however, it is only the x that has the power not the 8, so 8x^-1 = 8(1/x) = 8/x = 4 when x = 2
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  3. #18
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    Quote Originally Posted by novadragon849 View Post
    Hi thanks for the answers and I got most of them right but I got a question.

    On Q3 what does the graph suppose to look like?

    And on Q7a how did you get +4 on 8x^-1 to get the value for c. because if x = 2 I got 1/8(2) so I got 1/16 because I thought the power of -1 meant 1 over.

    Thank You
    8x^-1 does not equal (8x)^-1

    8x^-1 = 8/x
    (8x)^-1 = 1/(8x)

    Do you see the difference? In the first form, 8 is not being raised to the -1 power and so we do not take the reciprical of 8 when rewriting the problem.

    Therefore, at x = 2, we get
    8(2)^-1 = 8/2 = 4
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    for question 3. 1/x is a graph you should be very familiar with. adding 3 to it just shifts it up 3 units. here are the graphs, the red one is 1/x, the blue one is 1/x + 3
    Attached Thumbnails Attached Thumbnails Can anyone help me with these questions-inversex.jpg  
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  5. #20
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    Hi thanks I get the power thing now. in Q7a

    But if that is the case you wrote c= 1-8+12+4. Can you tell me how you got c=9 in the end cause I got 1=8-12 + 4 + c so c = 1-8+12-4 which I got c as 1

    Also has anyone managed Q9c yet cause i'm stuck
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  6. #21
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by novadragon849 View Post
    Hi thanks I get the power thing now. in Q7a

    But if that is the case you wrote c= 1-8+12+4. Can you tell me how you got c=9 in the end cause I got 1=8-12 + 4 + c so c = 1-8+12-4 which I got c as 1

    Also has anyone managed Q9c yet cause i'm stuck
    you are correct, it should be a - 4 not +4 so c = 1
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  7. #22
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    So on 3a I manage to understand on how to draw the graph but what are the equations for the asymtotes.
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by novadragon849 View Post
    So on 3a I manage to understand on how to draw the graph but what are the equations for the asymtotes.
    vertical asymptotes are lines of the form x = c
    horizontal asymptotes are lines of the form y = c

    so if you can find x and y values that are not possible for the function to attain, you found asymptotes.

    think of 1/x, are there any restrictions on x? you bet there are, x cannot be zero, or we are undefined, so guess what? x = 0 is a vertical asymptote, our graph cannot reach it.

    but this also affects y. because 1/x can never be zero with the restrictions we have on x. so y cannot be zero no matter what x we plug in, so y = 0 is a horizontal asymptote
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