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Math Help - The tagents line

  1. #1
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    The tagents line

    Hi all, I have a problem here and I don't know how to find a tangent line. Some hints may help me. Thank you so much.
    Let f(x) = x^2-x+3
    Find the straight lines that are tangent to the graph of f and going through the point (3/4,11/4)
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  2. #2
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    Re: The tagents line

    differentiate f(x) you will get d(f(x))=2x-1

    now substitute 3/4,11/4 in it as

    11/4=2(3/4)-1+c

    find c
    thus you will get a tangent line

    y=2x-1+c
    where c is the value which was found above
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  3. #3
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    Re: The tagents line

    I have the solution but I don't understand it. Can you explain it ?

    f'(x)=2x-1=(y-11/4)/(x-3/4)
    Therefore (2x -1)(x-3/4)=( x^2+x-3)-11/4 or x^2- 3/2 x + 1/2=0
    For x=1/2 the slope 2x-1=0 => the tangent line is y=11/4
    For x=1 the slope 2x-1=1 => the tangent line is y=x+2
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  4. #4
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    Re: The tagents line

    the last two points are obvious as (2x-1)=(y-11/4)/(x-3/4)
    puting x=1/2 in above will give y=11/4
    putting x=1 in above will give y=x+2
    and we have cross multiplied (x-3/4) with (2x-1) from above eqn
    so we get

    2x^(2)-(3/2)x-x+3/4=y-11/4

    now substituting y=x^(2)-x+3

    we get 2x^(2)-(3/2)x-x+3/4=x^(2)-x+3-11/4

    now solve and u will get

    x^(2)-(3/2)x+1/2

    now as slope of the function has to be found out on 0 so equate the above term with 0

    use breaking of middle terms so u split 3/2 into 1 and 1/2

    so u get

    (x-1/2)(x-1)=0

    so x=1/2 or x=1

    hence u get the above result
    Thanks from math88
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  5. #5
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    Re: The tagents line

    Quote Originally Posted by math88 View Post
    I have the solution but I don't understand it. Can you explain it ?

    f'(x)=2x-1=(y-11/4)/(x-3/4)
    Therefore (2x -1)(x-3/4)=( x^2+x-3)-11/4 or x^2- 3/2 x + 1/2=0
    For x=1/2 the slope 2x-1=0 => the tangent line is y=11/4
    For x=1 the slope 2x-1=1 => the tangent line is y=x+2
    $f(x) = x^2 - x + 3 \implies f'(x) = 2x - 1.$

    Let's say the tangent to f(x) when x = a (a is our unknown) runs through (3/4, 11/4). So the tangent's slope is 2a - 1.

    $\dfrac{\frac{11}{4} - (a^2 - a + 3)}{\frac{3}{4} - a} = 2a - 1 \implies\dfrac{4\left(\frac{11}{4} - a^2 + a - 3\right)}{4\left(\frac{3}{4} - a\right)} = 2a - 1 \implies \dfrac{11 - 4a^2 + 4a - 12}{3 - 4a} = 2a - 1 \implies$

    $- 1 - 4a^2 + 4a = 6a - 3 - 8a^2 + 4a \implies 4a^2 - 6a + 2 = 0 \implies a = \dfrac{- (-6) \pm \sqrt{(-6)^2 - 4 * 4 * 2}}{2 * 4} = \dfrac{6 \pm \sqrt{36 - 32}}{8} = 1\ or\ \dfrac{1}{2}.$

    At this point we have the x values for the points of tangency.

    Case I: $a = 1.$

    Slope = $2a - 1 = 2 * 1 - 1 = 2 - 1 = 1.$

    y value = $1^2 - 1 + 3 = 3$.

    Equation is $\dfrac{y - 3}{x - 1} = 1 \implies y - 3 = x - 1 \implies y = x + 2.$

    Check $y = x + 2 \implies y = \dfrac{3}{4} + 2 = \dfrac{11}{4}\ when\ x = \dfrac{3}{4}.$

    Case II: $a = \dfrac{1}{2}.$

    Slope = $2a - 1 = 2\left(\dfrac{1}{2}\right) - 1 = 1 - 1 = 0.$

    y value = $\left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 3 = \dfrac{1}{4} - \dfrac{1}{2} + 3 = 3 - \dfrac{1}{4} = \dfrac{11}{4}.$

    Equation is $\dfrac{y - \frac{11}{4}}{x - \frac{1}{2}} = 0 \implies y - \dfrac{11}{4} = 0 \implies y = \dfrac{11}{4}.$

    Check $y = \dfrac{11}{4} \implies y = \dfrac{11}{4}\ when\ x = \dfrac{3}{4}.$
    Last edited by JeffM; March 29th 2014 at 12:09 PM.
    Thanks from math88
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  6. #6
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    Re: The tagents line

    Why can't I solve the problem like this?
    The slope m=f’(x) =2x-1 =2. -1=1/2
    The tangent line is y-11/4=1/2(x-3/4 => y=1/2 x + 19/8
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  7. #7
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    Re: The tagents line

    Quote Originally Posted by math88 View Post
    Why can't I solve the problem like this?
    The slope m=f’(x) =2x-1 =2. -1=1/2
    The tangent line is y-11/4=1/2(x-3/4 => y=1/2 x + 19/8
    That would work IF the given point, (3/4, 11/4), were a point on the curve. But it isn't as you can see by setting x= 3/4 in x^2- x+ 3. The result is 45/16, not 11/4.

    You are asked to find the equation of a line that passes through a given point that is NOT on the curve but the line is tangent to the curve at some other point.
    Thanks from math88
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