posted by Kara .
Given the function defined as f(x)=x^3-(3/2)x^2-6x+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 x-axis
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
(x-3)(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^2-X-3. and then you should factor that to get x=-1 and x=2