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    The question asks,

    Find all tangents to the curve f( x)=x^2+3x+3 which go through the point (1,-2).
    This is what I've done so far,


    f'( x)=2x+3


    Gradient of tangest when x=1 is,


    m=2(1)+3


    m=5

    Equation of tangest is therefore,


    y-y_1=m(x-x_1)


    y-(-2)=5(x-1)


    y+2=5x-5


    y=5x-7


    That's the equation for one of the tangents, but I understand that there's another tangent to the curve, as the point (2,1) is 'below' the curve and as a result, there's a sort of a triangular shape which the two tangents create.

    How would I go about finding the other tangent?

    Thanks.
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    f'(x) is wrong
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    (Original post by volts)
    f'(x) is wrong
    Sorry, that was a typo. I've edited it now.
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    The point (1,-2) doesn't lie on f(x). I think you assume it does.
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    (Original post by volts)
    The point (1,-2) doesn't lie on f(x). I think you assume it does.
    The point (1,-2) lies on the tangents, not on the curve f( x).


    So in other wording, it's two tangents to the curve f( x) which intersect at the point (1,-2)
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    (Original post by hollywoodbudgie)
    The point (1,-2) lies on the tangents, not on the curve f( x).


    So in other wording, it's two tangents to the curve f( x) which intersect at the point (1,-2)
    So why have you evaluated f'(1)?
    Also, 5x-7 = f(x) has no real solution
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    (Original post by volts)
    So why have you evaluated f'(1)?
    Because I don't know how to work this out, hence I made this thread to ask for help.
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    (Original post by hollywoodbudgie)
    The point (1,-2) lies on the tangents, not on the curve f( x).


    So in other wording, it's two tangents to the curve f( x) which intersect at the point (1,-2)
    You know that the tangent passes through (1,-2). However, you also know that at the tangent crosses the parabola. So what I would do is find the possible coordinates which lies on the parabola. To do this note that the y coordinate at x1 is x1^+3*x1+3. Now using the formula for the equation of a line with only one point given you get an expression for the possible tangents. You know that the tangents must pass through 1,-2 so substitute these values in for x and y (not x1 and y1). Solve the resulting quadratic equation and you have found the x coordinates which satisfy the conditions given in the question. I'll leave the rest to you
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    (Original post by anshul95)
    note that the y coordinate at x1 is x1^+3*x1+3.
    Did that equation come out right, I don't particularly understand what the bolded symbol means, '^'. Would it be possible for you to write that out a little clearer please?

    Thanks.
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    The gradient of the tangent at a point (c, f(c)) is given by f'(c). So, the equation of a tangent that goes through the point (a,b) and has gradient f'(c) is y-b = f'(c)(x-a). But you also want the point (c, f(c)) to lie on the line, and so you must have f(c)-b = f'(c)(c-a). Solve this for c to find the x-values where the tangent of the curve goes through the point. [You have to work out what to use for a,b by the way.]
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    It's 2 tangents which happen to pass through (1, -2). Doesn't mean they touch the curve at this point. They can pass through (1, -2) yet touch the curve elsewhere.

    Tangent: y = mx + c; (1, -2) satisfies this
    => -2 = m + c => c = -2 - m
    => y = mx - (m+2)

    Since the tangents touch the curve:
    mx - (m+2) = x^2 + 3x + 3
    => x^2 + (3 - m)x + (5 + m) = 0

    Tangent => Discrim = 0
    => (3 - m)^2 - 4(m + 5) = 0
    => 9 - 6m + m^2 - 4m - 20 = 0
    => m^2 - 10m - 11 = 0
    => (m + 1)(m - 11) = 0
    => m = 11, m = -1

    Sub m = 11, m = - 1 into y = mx - (m+2) to get y = 11x - 13 and y = -x - 1.
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    (Original post by Physics Enemy)
    It's 2 tangents which happen to pass through (1, -2). Doesn't mean they touch the curve at this point. They can pass through (1, -2) yet touch the curve elsewhere.

    Tangent: y = mx + c; (1, -2) satisfies this
    => -2 = m + c => c = -2 - m
    => y = mx - (m+2)

    Since the tangents touch the curve:
    mx - (m+2) = x^2 + 3x + 3
    => x^2 + (3 - m)x + (5 + m) = 0

    Tangent => Discrim = 0
    => (3 - m)^2 - 4(m + 5) = 0
    => 9 - 6m + m^2 - 4m - 20 = 0
    => m^2 - 10m - 11 = 0
    => (m + 1)(m - 11) = 0
    => m = 11, m = -1

    Sub m = 11, m = - 1 into y = mx - (m+2) to get y = 11x - 13 and y = -x - 1.
    Thank you
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    (Original post by hollywoodbudgie)
    Thank you
    No worries. If it's any consolation, I didn't have a clue what some of the other posters were on about either. Not sure why people make Maths harder than it has to be; it's already hard enough by itself.
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    (Original post by Physics Enemy)
    No worries. If it's any consolation, I didn't have a clue what some of the other posters were on about either. Not sure why people make Maths harder than it has to be; it's already hard enough by itself.
    This relates to me more than you could possibly imagine. :wink2:
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    (Original post by hollywoodbudgie)
    This relates to me more than you could possibly imagine. :wink2:
    Sorry if we confused you. Some of us just did it another way.
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    (Original post by anshul95)
    Sorry if we confused you. Some of us just did it another way.
    :hugs: Thanks for the help.
 
 
 
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