Posted by **Kara** on Thursday, November 10, 2011 at 10:43pm.

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 !

- Calculus -
**Damon**, Friday, November 11, 2011 at 5:31am
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.

- Calculus -
**Anonymous**, Wednesday, January 25, 2012 at 5:05pm
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

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