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Homework Help: Intersection point between a line with slope m and f(x)

  1. Jan 11, 2010 #1

    cir

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    1. The problem statement, all variables and given/known data
    f(x) = 2x2-5x-12 is given;
    part a: find derivative of f(x) using first principles,
    part b: find the rate of change of f(x) at x=1,
    part c: the points at which the line through (1, -15) with slope m cuts the graph of f(x),
    part d: the value of m such that the points of intersection found in (c) are coincident,
    part e: the equation of the tangent to the graph of f(x) at the point (1, -15)

    2. The attempt at a solution
    I found the derivative first, using first principles, it is f'(x) = 4x - 5

    Then for part b, the rate of change of f(x) at x = 1, it's the f'(x) value at x = 1, which is -1, the slope of tangent.

    I am stuck at part c. I first got the line equation; y-y1=m(x-x1) using point (1, -15), which gives y = mx-m-15

    So to find the intersection points with this line and f(x), I tried to equate them both and solve,

    mx-m-15 = 2x2-5x-12, and end up at the following,

    2x2-(5-m)x+(m+3)=0 how do I solve this? or is my approach wrong?

    And what is that part d is asking? I didn't understand it.

    Thanks!
     
  2. jcsd
  3. Jan 11, 2010 #2

    rock.freak667

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    Homework Helper

    You are correct thus far.

    Part d is saying that the equation you found 2x2-(5-m)x+(m+3)=0 has only one root.
     
  4. Jan 11, 2010 #3

    rl.bhat

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    Hi cir, welcome to PF.
    2x2-(5-m)x+(m+3)=0
    This equation should be
    2x2-(5+m)x+(m+3)=0
    If ax^2 + bx + c = 0, then x = [-b (+ or -)(b^2 - 4ac)^1/2 ]/2a
    Use this formula and solve for x.
     
  5. Jan 11, 2010 #4

    cir

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    @rl.bhat

    Thanks. Just saw the sign error. Now able to solve using factorization too.

    mx-m-15=2x2-5x-12

    2x2-5x-mx+m+3=0

    2x2-5x+3-m(x-1)=0

    2x(x-1)-3(x-1)-m(x-1)=0

    (2x-3-m)(x-1)=0

    x = 1 ; x = (m+3)/2

    y= -15 ; y = ((m2+m)/2)-15

    intersecting points therefore: pt (1, -15), pt ((m+3)/2, ((m2+m)/2)-15) ;answer correct according to text book answers.


    @rock.freak667

    I still don't get it :frown: , now that I have found the intersection points, how do I find the value of m such that the points of intersection found in (c) are coincident? how do I know it's just one root (cause I did get two intersection points, (1, -15) and the other in terms of m)?

    My guess is since the pt (1, -15) is on both lines, thus coincident to both lines, I could equate the x values of the other intersection point found in terms of slope m ((m+3)/2, (m2+m)/2-15)

    Thus equating x coordinates gives (m+3)/2 = 1 and solving for m, m = -1 (equating y coordinates gives m = 0)

    m = -1 is the correct answer in book, but I still don't understand this part completely. some further explanation would be great :smile:
     
  6. Jan 11, 2010 #5

    rl.bhat

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    The tangent to the curve touches the curve at one point.
    You have already found f'(x) = m = (4x - 5). Put x = 1. You get m = -1.
     
  7. Jan 11, 2010 #6

    cir

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    got it, thnks.
     
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