Calculus

posted by Anonymous

1. Given the function f defined by f(x) = x^3-x^2-4X+4
a. Find the zeros of f
b. Write an equation of the line tangent to the graph of f at x = -1
c. The point (a, b) is on the graph of f and the line tangent to the graph at (a, b) passes through the point (0, -8) which is not on the graph of f. Find the values of a and b.

I'm positive I can do a and b, but what about c?

  1. Mgraph

    a)f(x)=(x+2)(x-1)(x-2)

    c)From the graph a>=2, b>=0.
    The equation of the tangent
    y=b+(3a^2-2a-4)(x-a) and if x=0
    y=b-(3a^3-2a^2-4a) or
    y=a^3-a^2-4a+4-(3a^3-2a^2-4a)
    y=-2a^3+a^2+4=(2-a)(6+3a+2a^2)-8

    If a>2 then y<-8

    (a,b)=(2,0)

  2. Reiny

    inherently the same as Mgraph's, but presented in slightly different way

    dy/dx = f'(x) = 3x^2 - 2x - 4
    at (a,b)
    dy/dx = 3a^2 - 2a-4 , which is the slope of the tangent at (a,b)
    but the slope of the tangent passing through (a,b) and (0,-8) is also
    = (b+8)(a-0) = (b+8)/a

    so (b+8)/a = 3a^2 - 2a - 4
    b+8 = 3a^3 - 2a^2 - 4a
    b = 3a^3 - 2a^2 - 4a - 8
    but b = a^3 - a^2 + 4a + 4

    3a^3 - 2a^2 - 4a - 8 = a^3 - a^2 + 4a + 4
    2a^3 - a^2 - 8a - 4 = 0
    a^2(a-2) - 4(a-2) = 0
    (a-2)(a^2 - 4) = 0
    (a-2)(a-2)(a+2) = 0
    a = 2 or a = -2

    but from a = -2 we cannot have a tangent to (0,-8)
    so if
    a = 2, then b = 8 - 4 - 8 + 4 = 0

    (a,b) = (2,0)

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