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calculus

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find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=-2 to solve the line with slope-4 and tangent to the line y=-8

  • calculus - ,

    Start with defining the function:
    f(x)=ax²+bx+c
    Find its derivative:
    f'(x)=2ax+b
    1. f'(1)=8 =>
    2a(1)+b = 8 ....(1)
    2. f'(-2)=-4 =>
    2a(-2)+b = -4 ...(2)
    Solve system of equations (1) and (2) for a and b. (a=2,b=4)

    The parabola is tangent to the line y=-8 => y=-8 is the minimum.

    Now find the minimum of f(x) by setting f'(x)=0 and solve for x.
    f'(x)=2ax+b=0
    2(2)x+(4)=0
    x=-1
    So
    f(-1)=-8
    2(-1)²+4(-1)+c = -8
    so c=-6
    =>
    f(x) = 2x²+4x-6

    Do some checking to make sure the function satisfies all the conditions (and in case I make an arithmetic error):
    given conditions to be checked:
    f(1)=8
    f(-2)=-4
    minimum f(x)=-8

  • calculus - ,

    find the slope of the tangent line to the graph of e^xy=x at (1,0)

    write an equation of the tangent line

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