calculus

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Determine the equations of two lines that pass through the point (1, -5) and are tangent to the graph of y = x^2 - 2.

  • calculus -

    y=-2x-3, y=6x-11

  • calculus -

    We are given a fixed point through which the tangents pass.

    On the curve (parabola), the slope of the tangent is dy/dx = 2x at the point (x,x²-2).

    What we need here is a line that passes through a point (x,f(x)) and (1,-5) with a slope of 2x.

    Thus we form the equation:
    (y2-y1)/(x2-x1)=slope
    (x²-2 - (-5))/(x-1) = 2x
    We solve for x=3 or x=-1.
    Substitute into equation
    (y-y0)=m(x-x0)
    to get the formulae alx posted.

  • calculus -

    y = x^2 - 2

    f' = 2x
    P(1,-5)
    f' = slope m = 2(1) = 2

    Equation of the line tangent at P(1,-5)
    m = 2
    y = mx + b
    -5 = 2(1) + b
    -5 = 2 + b
    b = -7

    Equation of the tangent line is,
    y = 2x - 7
    2x - y = 7

    Equation of the normal line at P(1,-5) (The normal line is the line that is perpendicular to the tangent line at the point of tangency).

    m = -1/2
    y = mx + b
    -5 = -1/2(1) + b
    -5 = -1/2 + b
    b = -10/2 + 1/2
    b = -9/2

    Equation of the normal line
    y = -1/2 x - 9/2
    2y = -x - 9
    x + 2y = -9

  • calculus -

    I think I read the question wrong.

    I thought the question wanted the slope of the tangent of f(x) at (1,-5) and the two equations at this point.

  • calculus -

    Ignore my response.

    I completely misunderstood the question.(:

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