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Math Help - Tangent to a curve

  1. #1
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    Tangent to a curve

    Consider the function f(x)= x^2+k, k\in \mathbb{R} and the tangent to the curve at x=a

    Show that the tangent does not pass through the origin if k<0 .

    It seems obvious that it does that it pass through the origin when I draw the function for k<0 as the origin is 'inside' the parabola. But is this argument enough?

    I also have the equation of the tangent being \displaystyle y = f'(a)x+c \implies y = 2ax+c
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  2. #2
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    Re: Tangent to a curve

    Quote Originally Posted by Bushy View Post
    Consider the function f(x)= x^2+k, k\in \mathbb{R} and the tangent to the curve at x=a

    Show that the tangent does not pass through the origin if k<0 .

    It seems obvious that it does that it pass through the origin when I draw the function for k<0 as the origin is 'inside' the parabola. But is this argument enough?

    I also have the equation of the tangent being \displaystyle y = f'(a)x+c \implies y = 2ax+c
    you need to show the y-intercept of the tangent line cannot equal zero.

    f(x) = x^2+k \, ; \, k < 0

    f'(x) = 2x

    at the point on the curve (a, a^2+k) , the tangent line equation is

    y - (a^2+k) = 2a(x - a)

    y = 2ax - 2a^2 + (a^2 + k)

    y = 2ax - (a^2 - k)

    k < 0 \implies (a^2 - k) \ne 0
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  3. #3
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    Re: Tangent to a curve

    An alternative solution:


    At x = a, the tangent line goes through the point (a, a^2 + k) and the slope is 2a. Therefore, writing the equation of the line in slope-intercept form,

    a^2 + k = 2a(a) + b, where b is the y-intercept.

    k = a^2 + b. If b = 0, then we get k = a^2, which cannot be true since k < 0 and the LHS is always non-negative.
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