# Can anyone help me with these questions

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• Apr 25th 2007, 11:15 PM
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
• Apr 25th 2007, 11:18 PM
Jhevon
Quote:

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
• Apr 25th 2007, 11:21 PM
ecMathGeek
Quote:

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
• Apr 25th 2007, 11:21 PM
Jhevon
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
• Apr 26th 2007, 12:45 AM
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
• Apr 26th 2007, 12:54 AM
Jhevon
Quote:

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
• Apr 26th 2007, 01:06 AM
So on 3a I manage to understand on how to draw the graph but what are the equations for the asymtotes.
• Apr 26th 2007, 01:16 AM
Jhevon
Quote: