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Pesky Tangents

  1. Mar 7, 2006 #1
    Ok, I've found the answer to these questions, but I did so in more of a trial and error way. The question is:

    f(x) = 2x^2 + x

    Find the two tangent lines to the curve, which both pass through the point (2,-3).

    So, I tried using y=mx+b = 2x^2 + x, where m = f'(x) = 2x + 1, thus:
    (2x+1)x + b = 2x^2 + x. I worked this out to find an extraneous answer, and one correct tangent of y = -x - 1. I know the other answer has a slope of 11, through trial and error, but there's got to be an easier way.

    Any thoughts on how to do this algebraically?
     
  2. jcsd
  3. Mar 7, 2006 #2

    benorin

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    Ok, so [tex]f(x) = 2x^2 + x\Rightarrow f^{\prime} (x) = 4x + 1[/tex]

    We wish to find the two tangent lines to the curve, which both pass through the point (2,-3). Use the point-slope formula instead, that is, use

    [tex]y-y_0=m(x-x_0)[/tex]

    Since the tangent line(s) pass throught the point (2,-3), we may take [tex](x_0,y_0)=(2,-3),[/tex] hence

    [tex]y+3=m(x-2)[/tex]

    but also, if (a,f(a)) is the point at which the line is tangent to the curve, we require that, [tex]m=f^{\prime} (a) = 4a + 1[/tex] and that (a,f(a)) be a point on the tangent line itself, hence we require that

    [tex]f(a)+3=f^{\prime} (a)(a-2) \Rightarrow 2a^2+a+3=(4a+1)(a-2) [/tex]

    hold. The two solutions for a give the two tangent lines.
     
  4. Mar 7, 2006 #3
    Ahh, I see. It looks like I wrote the function down wrong (should be x^2 + x), but thanks for the insight.
     
  5. Mar 7, 2006 #4

    benorin

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    That you may check your work: the values of a that work for f(x)=x^2+x are [tex]a=2\pm 3 = -1,5[/tex] and the other tangent line is y=11x-25
     
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