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Math Help - parabola tangent

  1. #1
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    parabola tangent

    a straight line passing through points (1,-2) is tangent of the parabola y=x^2, find the equation of the straight line.

    and also what is meant by the term 'linear factors'? please explain it in general and in terms of 12x^2 + 7x - 12

    and also explain how this works y = 2(2-x)^2 + 3 i got up to: -2(x-2)^2 + 3

    the book however sketched the graph without it being negative.

    Thanks alot
    Last edited by andrew2322; February 25th 2008 at 02:12 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by andrew2322 View Post
    and also what is meant by the term 'linear factors'? please explain it in general and in terms of 12x^2 + 7x - 12
    when do you say if a polynomial is linear? so, linear factors are those factors which are linear polynomials..
    for example, if you have x^2 + 5x + 6, then the linear factors are x + 3 and x + 2..
    that is, x^2 + 5x + 6 = (x+3)(x+2)..

    Quote Originally Posted by andrew2322 View Post
    and also explain how this works y = 2(2-x)^2 + 3 i got up to: -2(x-2)^2 + 3

    the book however sketched the graph without it being negative.

    Thanks alot
    no.. take note that if you factor the negative there, (2-x)^2, you should have (-(x-2))^2 = (-1)^2(x-2)^2 and hence (x-2)^2, that is, there is no negative sign..
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  3. #3
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    hey

    hey kalagota, thanks for the reply however i still dont understand how it becomes one? how come u just took the negative out and put them in brackets with alone, how come it didnt end up with the 2?

    consider this example 400(x-x^2)
    = 400x(1-x)

    in this case the x was factored out and just put infront of the 400

    thanks alot.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by andrew2322 View Post
    a straight line passing through points (1,-2) is tangent of the parabola y=x^2, find the equation of the straight line.
    i suppose you want the non-calculus way of doing this

    there will be two solutions here. see the graph below.

    Let the line be y = mx + b

    since it passes through (1,-2), we know that -2 = m + b ...........(1)

    Also, since the line is tangent to y = x^2, it means they intersect at one point. thus, there is one solution to

    x^2 = mx + b

    \Rightarrow x^2 - mx - b = 0

    by the quadratic formula:

    \Rightarrow x = \frac {m \pm \sqrt{m^2 + 4b}}2

    since we want one root, we must have m^2 + 4b = 0 ...............(2)


    Now, by (1), b = -m - 2, plug this into (2), we get:

    m^2 + 4(-m - 2) = 0

    \Rightarrow m^2 - 4m - 8 = 0

    again, by the quadratic formua:

    m = \frac {4 \pm \sqrt{48}}2

    \Rightarrow \boxed{m = 2 \pm \sqrt{12}}


    if m = 2 + \sqrt{12}, then b = -4 - \sqrt{12}

    so, our line is: \boxed{y = (2 + \sqrt{12})x + (-4 - \sqrt{12})}

    if m = 2 - \sqrt{12}, then b = \sqrt{12} - 4

    so, our line is: \boxed{y = (2 - \sqrt{12})x + (\sqrt{12} - 4)}

    taking either of these (ugly) lines work
    Attached Thumbnails Attached Thumbnails parabola tangent-tangentto.jpg  
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by andrew2322 View Post
    hey kalagota, thanks for the reply however i still dont understand how it becomes one? how come u just took the negative out and put them in brackets with alone, how come it didnt end up with the 2?

    consider this example 400(x-x^2)
    = 400x(1-x)

    in this case the x was factored out and just put infront of the 400

    thanks alot.
    what is the square of negative numbers? in particluar, what is (-1)^2?

    note that, (ab)^n = a^nb^n..

    in your example, 400(x-x^2) = 400x(1-x) and if you take out the negative, it will become 400x(-1)(x-1) = -400x(x-1)

    considering your problem, y = 2(2-x)^2 + 3 = 2(-(x-2))^2 + 3 = 2(-1)^2(x-2)^2 + 3 and since the square of negative 1 is 1, therefore, you will get what the book says..
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