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Curve Equation

The question is from the gcse edexcel june 2008 calculator paper (q.25)

The sketch shows a curve with the equation y=ka^x where k and a are constants, and a > 0
The curve passes through the points (1,7) and (3, 175)
Calculate the value of k and the value of a....

Please help me!
Reply 1
In the co-ordinate (1, 7) 1 is your x-coordinate and 7 is your y-coordinate

if y = ka^x

then 7 = ka^1

same for the other point.

Then you have two equations which you can solve.
Reply 2
Substitute the values of x and y in and solve simultaneously.
Reply 3
So going by your methodology this should eventually turn into simultaneous equations like this:
7= ka^1
175=ka^3
what would you do next though :s
Reply 4
Nandoz
So going by your methodology this should eventually turn into simultaneous equations like this:
7= ka^1
175=ka^3
what would you do next though :s


7=ka    a=7k7 = ka \implies a = \frac{7}{k}

You can use this fact to write the second equation in terms of just k.
Reply 5
Rocious
7=ka    a=7k7 = ka \implies a = \frac{7}{k}

You can use this fact to write the second equation in terms of just k.


ohh right...so you get 175= k (7/k)^3 but then how do you rearrange to get a value for k? sorry for not getting this
Reply 6
would you get 175= 343k^2?
Reply 7
175=k(7k)3    175=k(73k3)    175=343k2175 = k(\frac{7}{k})^3 \implies 175 = k(\frac{7^3}{k^3}) \implies 175 = \frac{343}{k^2}
Reply 8
Rocious
175=k(7k)3    175=k(73k3)    175=343k2175 = k(\frac{7}{k})^3 \implies 175 = k(\frac{7^3}{k^3}) \implies 175 = \frac{343}{k^2}



oh so k= sqroot (343/175)?
Reply 9
Nandoz
oh so k= sqroot (343/175)?


Well yeah.

BUT

there are 2 solutions to that equation.

However when you try to find a you realise that one of those isn't valid, so k = sqrt(343,175) is correct.
Reply 10
Thank you Rocious- rep added
Couldn't you have done k/7 and then 245/7=35. If that's incorrect then is A simply 7/k?
Reply 12
Original post by Sophiemacyds
Couldn't you have done k/7 and then 245/7=35. If that's incorrect then is A simply 7/k?


putting the content aside, have you noticed the date when this question was debated?
Original post by TeeEm
putting the content aside, have you noticed the date when this question was debated?

I just saw the date - I was looking for an answer as I needed help on this question on my summer maths homework. Was the content correct?
Reply 14
Original post by Sophiemacyds
I just saw the date - I was looking for an answer as I needed help on this question on my summer maths homework. Was the content correct?


I did not look at the conversation
My advice is to post a new thread with a photo/scan of the question and I am sure they will be some helpers soon
Original post by TeeEm
I did not look at the conversation
My advice is to post a new thread with a photo/scan of the question and I am sure they will be some helpers soon

Okay - thanks! :smile:
Wow 14 and 7 years after these people and now I'm doing this question!

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