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a) determine the equation of the line that is perpendicular to the tangent to y=5x^2 at (1,5).

b) determine the equation of the line that passes through (2,2) and is parallel to the line tangent to y=-3x^3-2x at (-1,5).

  • Math -

    (a)
    y=5x²
    dy/dx=5*2x=10x, slope of tangent
    slope of line perpendicular to tangent
    = -1/(10x)
    At (1,5), slope of perpendicular line
    = -1/(10*1) = -1/10
    Line passing through (1,5)
    (y-5)=-(1/10)(x-1)
    y =-(1/10)x + 5.10

    Test for perpendicularity at (1,5)
    -(1/10) * (10*1)
    = -1 Line and tangent are perpendicular.

    (b) y=-3x^3-2x at (-1,5).
    dy/dx = -9x² -2 = -9-2 = -11 at (-1,5)
    Slope of tangent = -11
    Line passing through (2,2) with slope of -11
    (y-2) = -11(x-2)
    y-2 = -11x + 22
    y = -11x + 24

  • Math -

    slope of -8y-3x+5=10x

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