Posted by Anonymous on Monday, January 16, 2012 at 12:52am.
1. Given the function f defined by f(x) = x^3x^24X+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?

Calculus  Mgraph, Monday, January 16, 2012 at 2:16am
a)f(x)=(x+2)(x1)(x2)
c)From the graph a>=2, b>=0.
The equation of the tangent
y=b+(3a^22a4)(xa) and if x=0
y=b(3a^32a^24a) or
y=a^3a^24a+4(3a^32a^24a)
y=2a^3+a^2+4=(2a)(6+3a+2a^2)8
If a>2 then y<8
(a,b)=(2,0)

c)  Calculus  Reiny, Monday, January 16, 2012 at 9:58am
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  2a4 , 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)(a0) = (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(a2)  4(a2) = 0
(a2)(a^2  4) = 0
(a2)(a2)(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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