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# cant get the answer!! watch

1. ok im just doing some practice for maths on simultaneous equations. I bought this book the others day and i really can get the answer that they get in the back of the book. The question is:
Find the possible values of k f y=2x+k meets y=x^2-2x-7
I tried to alter the 2nd equation to x^2-2x-7-y and then put the first equation in but i think its wrong. Any help?
The answer is K>-11 if that helps.
Thanks xxxxx
2. hi!
i dunno if someone has put the answer up while i'm writing this...but i'll write it anyway

the way i've done it is this:
equate the 2 equations to give 2x+k = x^2-2x-7
which then simplifies to: x^2-4x-7-k = 0
now...complete the square to give: (x-2)^2 -11-k = 0
bring the k over to the other side to give (x-2)^2-11 = k
now...look at the let hand side and you can see that the minimum point occurs at -11
hence, k must be >-11

hope this makes sense!
3. (Original post by gobbledygook88)
hi!
i dunno if someone has put the answer up while i'm writing this...but i'll write it anyway

the way i've done it is this:
equate the 2 equations to give 2x+k = x^2-2x-7
which then simplifies to: x^2-4x-7-k = 0
now...complete the square to give: (x-2)^2 -11-k = 0
bring the k over to the other side to give (x-2)^2-11 = k
now...look at the let hand side and you can see that the minimum point occurs at -11
hence, k must be >-11

hope this makes sense!
hey, thanks that helped. But i just have one question, how do you know from looking at (x-2)^2-11 that the minimum point is -11? Oh and if the questions says in just one point, how do you do that?
xxxxx
4. (Original post by indie_couture)
how do you know from looking at (x-2)^2-11 that the minimum point is -11?
Think of the graph

It is a U shaped parabola with a minimum at (0,0).

Now, if you understand graph transformations, you should see that the graph of is also a U shaped parabola but it has been translated 2 units in the x direction and -11 units in the y direction so the minimum point is now (2, -11).
5. you can see where the min/max point of a graph is by looking at (x-2)^2-11 by using transformations...in this case its only translations [ie moving the graph in the x and y directions on a single plane]

starting with y=x^2 [a parabola with min vertex at origin]
translate 2 units to the right to give y=(x-2)^2 [min vertex at (2,0)]
translate 11 units down to give y=(x-2)^2-11 [min vertex at (2,-11)]
this then shows that the min point of the graph is -11

erm...about the other part...i dont actually know what that means...could you write out the whole question in full?
6. erm we actually havent gone over graphs of transformations yet, although i remember bits of what you said from GCSE.
The question is
Find the possible values of k f y=2x+k meets y=x^2-2x-7 then part b just says in just one point. ??
xxxxxxx
7. ah...i think i have it...
from the working above, you can also see that for the line y=2x+k to touch the curve y=x^2-2x-7 once, the value of k=-11

you can check this by drawing a graph with both equations
or by differentiating both original equations, find the x-coordinate where the gradients of both equations are the same, find the corresponding y-coordinate
and then solve, using y=2x+k, to find k, which also gives k=-11

i hope its right...
8. (Original post by gobbledygook88)
ah...i think i have it...
from the working above, you can also see that for the line y=2x+k to touch the curve y=x^2-2x-7 once, the value of k=-11

you can check this by drawing a graph with both equations
or by differentiating both original equations, find the x-coordinate where the gradients of both equations are the same, find the corresponding y-coordinate
and then solve, using y=2x+k, to find k, which also gives k=-11

i hope its right...
yeh thats right
thankyou so much, youve helped a lot!
what exam board are you on??
xxxxx
9. i'm on OCR
fun fun fun!!!!
send over a message with any more questions whenever you have problems...happy to help...also gives me a chance to refresh my memory of last year
10. (Original post by gobbledygook88)
i'm on OCR
fun fun fun!!!!
send over a message with any more questions whenever you have problems...happy to help...also gives me a chance to refresh my memory of last year
thanks, i will
We've been doing some stuff on geometry which some parts im struggling with, but i'm going to over it tomorrow, so might msg you then.
Once again thanks xxxx

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