Calculus
posted by Kara on .
Given the function defined as f(x)=x^3(3/2)x^26x+10
a) Explain why f(x) must have a root between x=3 and x=2
b) Write an equation of the line perpendicular to the graph of f at x=0
c) Find the x and y coordinates of the point on the graph of f where the line tangent to the graph is parallel to the xaxis
Help would be GREATLY appreciated !

The value of the function at x = 3 is
27 27/2+18+10 = 12.5
at x = 2 it is
8 12/2 +12+10 = 8
the function is continuous so must pass through zero on the way from 12.5 to +8
dy/dx = 3 x^2 3x 6
at x = 0, dy/dx = 6
so slope of our line = +1/6
at x = 0, y = 10
so find equation of the line that passes through (0,10) with slope m = 1/6
dy/dx = 0 = 3 x^2  3x 6
x^2  x  6 = 0
(x3)(x+2) = 0
so parallel to x axis at x = +3 and x = 2
find y at those points. 
That is actually incorrect for the last part. The equation should turn out to be x^2X3. and then you should factor that to get x=1 and x=2