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    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
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    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!
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    (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
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    (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 y=x^2

    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 y=(x-2)^2-11 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).
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    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?
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    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
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    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
    is this the answer?

    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...
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    (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
    is this the answer?

    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
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    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
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    (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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