Question 388614
2a.  A(2,0) says that y=0 when x=2.
To solve for k, substitute 2 for x and 0 for y in the given expression, y=x^3-Kx^2-16x+32
0=2^3-2^2K-16(2)+32
4k =8-32+32
4k = 8
k=2

2b. B(p,35) says that when x=p in the given expression, y = 35
To solve for p, substitute p for x, y =35, in the same expression
35 =p^3-(2)p^2-16p +32
0 = p^3-2p^2-16p-3

To solve for p, we can use the possible rational roots theory or, if allowed, use the graphing calculator to determine the roots. According to the rational roots theory, the only possible rational roots are (+ or -1) and (+or-3).   If p=-3, the expression becomes 3^3-2(3^2)-16(-3)-3=-27-18+48-3 = 0, and the equation is satisfied.  The other rational roots will not satisfy the  equation. In addition to -3, p is also equal to two other irrational roots.  They can be obtained by dividing the original third degree expression by (x+3) or use synthetic division which will give a second degree equation from which can be determined the two irrational roots.

The second degree equation obtained from division is x^2-5x -1
using the quadradic formula, x = (5 + or - Sqr root of 29)/2

Using the rational root of -3 only, A(2,0) and B(-3,35) the angle between the line AB and the x-axis in the  positive direction is determined as follows:

As can be shown on a graph, line AB is the hypotenuse of a right triangle with legs of 35 opposite and 5 adjacent to a supplement of the angle in question. 

arc tan (35/5) = 81.87 deg (this is the angle AB makes with the x-axis in the negative direction)
for the positive direction, the angle = 180-81.87 = 98.13 deg


ans: k =2
p=3. or( 5+ or - sqr root of 29)/2 (3 roots, 1 rational and 2 irrational)
Angle line AB makes with x-axis in the positive direction (using (-3,35) point only) = 98.13 deg