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Math Help - Find values of a and b...

  1. #1
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    Find values of a and b...

    I'm really having trouble with these two problems in my homework... I'm just not sure where to start and would appreciate a push in the right direction.

    In my class, we've just gotten through the quotient rule for derivatives.

    1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2


    This is what I did, but my answer did not check out when I graphed it and such:

    a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?
    So...
    a + b = 1 or b = 1-a or a = 1-b

    y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
    2a(1) + b = 3
    So...
    2a + b = 3

    2a + (1-a) = 3
    a = 4

    2(1-b) + b = 3
    2-b = 3
    b = -1

    But when I graphed both, the equation of the tangent line didn't lie on (1,1) of the graph y = 4x^2 - x. Where did I go wrong?

    2. Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y= -x^2 +6x - 5.

    And I don't know where to start for this one...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    I'm really having trouble with these two problems in my homework... I'm just not sure where to start and would appreciate a push in the right direction.

    In my class, we've just gotten through the quotient rule for derivatives.

    1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2


    This is what I did, but my answer did not check out when I graphed it and such:

    a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?
    So...
    a + b = 1 or b = 1-a or a = 1-b

    y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
    2a(1) + b = 3
    So...
    2a + b = 3

    2a + (1-a) = 3
    a = 4
    your mistake was in this last line. you should subtract 1 from both sides, not add it. a = 2
    Last edited by Jhevon; September 24th 2007 at 10:15 PM.
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  3. #3
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    Thanks, I knew I did something ridiculous... can't believe I didn't catch that one while typing.
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  4. #4
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    1. Find values of a and b so that y = ax^2 + bx has a tangent line at (1,1) whose equation is y = 3x - 2

    a(1)^2 + b(1) = 3(1) - 2 <---This is a guess, can I do this?

    The Lefthand side is okay, but the RHS is...where did you get that.

    The RHS should be 1 only.
    Yours came out as 1 also, but that is by chance only.

    y = ax^2 +bx
    At (1,1),
    1 = a(1^2) +b(1)
    1 = a +b ----------------------(1)

    y' = 2ax + b =3 <---since the slope of the tangent line is 3, at (1,1)
    2a(1) + b = 3
    So...
    2a + b = 3 -------------------(2)
    Since, from (1), b = 1-a, then, in (2),
    2a +(1-a) = 3
    2a -a = 3 -1
    a = 2 ----------------------------------answer.
    And so, b = 1 -2 = -1 ----------------answer.

    Check,
    y = ax^2 +bx
    y = 2x^2 +(-1)x
    y = 2x^2 -x
    Is (1,1) a point of the curve?
    1 =? 2(1^2) -1
    1 =? 1
    Yes, so, OK.

    ---------------------------------------------------------

    2. Find the equation of the line(s) that are tangent to the graphs of y=x^2 and y= -x^2 +6x - 5.

    Say there is only one common line that is tangent to both curves

    On the y = x^2 parabola,
    m1 = y' = 2x

    On the y = -x^2 +6x -5 parabola,
    m2 = y' = -2x +6

    It is the same tangent line, so its slope is the same everywhere along the line.
    So, m1 = m2
    2x = -2x +6
    4x = 6
    x = 6/4 = 3/2
    And so,
    m1 = 2x = 2(3/2) = 3
    m2 = -2x +6 = -2(3/2) +6 = 3 also.

    So the equation of the tangent line is y = 3x +b -----(i)

    (i) intersects the 1st parabola at their point of tangency, so, their y's there are the same. Hence,
    3x +b = x^2
    x^2 -3x -b = 0 -------------------(ii)

    Likewise, on the 2nd parabola,
    3x +b = -x^2 +6x -5
    x^2 -6x +3x +b +5 = 0
    x^2 -3x +5 +b = 0 ---------------(iii)

    (iii) minus (ii),
    5 +b +b = 0
    2b = -5
    b = -5/2 --------------***

    Therefore, the equation of the tangent line is
    y = 3x +b -----(i)
    y = 3x -5/2 ---------------------answer.
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