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Equation of tangent questions

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the function
    y = x[tex]^{2}[/tex] + 3x - 4 find,

    a) the equation of tangent + normal at x = -2.
    b)the equation of the tangent that has slope of -2.
    c)local min and max.
    d)the equations of the tangent from the point (2 , -4) to the curve.
    e) area enclosed by the tangent at x = -2 and the axis.

    3. The attempt at a solution
    ok so i did parts a to c. Im kind of stuck at part d and e. I dont understand what the questions mean.
  2. jcsd
  3. Jan 12, 2009 #2
    d) Notice that (2, -4) is not on the curve. We want the equations of all lines passing through (2, -4) that are tangent to the curve. There could be 0, 1, or 2 of these lines because y is a quadratic. Let (x, y) be a point on one such line that's also on the curve. Then you know what y is in terms of x, and you can find the slope of the line in 2 ways: by differentiation, and by rise over run from (x, y) to (2, -4).

    e) You found the equation of the tangent at x = -2 in part (a). Graph it and calculate the area of the triangle enclosed by this line, the x-axis, and the y-axis.
  4. Jan 12, 2009 #3


    Staff: Mentor

    Part d is a bit tricky, I think. You want to find a point A on the parabola with coordinates (p, p^2 + 3p - 4) that satisifies these conditions:
    1. The slope of the segment from A to (2, -4) is equal to the slope of the tangent at A (p, p^2 + 3p -4).
    2. There is only one value of p that works.
    I haven't worked through this, but this is the approach I would take for starters. The reason for the second condition is that the tangent has to touch the graph of the parabola at exactly one point; it can't intersect the graph at a point and then continue to intersect the other side of the parabola graph.

    Part e is not well-defined. Did the original problem say "area enclosed by the tangent at x = -2 and the axes"? One axis, two axes. If so, the area involved is a triangle, so if you can find the intercepts of the tangent on the two axes, it's pretty easy to find the area of that triangle.
  5. Jan 12, 2009 #4
    I dont understand why the tangent slope is equal to the regular slope.
  6. Jan 12, 2009 #5


    Staff: Mentor

    I'm not sure what you're asking. What do you mean by "regular slope"?
  7. Jan 13, 2009 #6
    sorry. I was referring to the slope from point A to point (2 , 4)
  8. Jan 13, 2009 #7


    Staff: Mentor

    Your problem asks for "the equations of the tangent from the point (2 , -4) to the curve."
    I interpret this to mean the line that is tangent to the given curve at A(p, p^2 + 3p + 4)and that passes through (2, -4). Can you find the slope of this line segment? Can you find the slope of the tangent line to the curve at point A? The two expressions you get will involve p and have to be equal.
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