calc 150 -- due in 5 min!!?

Question

A straight line, with positive slope m , passes through the origin. The curve y=x(x–1)(2–x) also passes through the origin. For how many values of m , if any, is the straight line a tangent to the curve?

Answer 1

curve
y=x(x-1)(2-x) =x( -x^2 +3x -2) = -x^3 +3x^2 -2x

line equation
y=mx+b
(0,0) is on the line so
0=m *0 +b
b=0
so
y=mx

the slope is tangent to the curve so
m=y' = -3x^2 +6x-2
so
y=mx
y=(-3x^2 +6x -2) x =-3x^3+6x^2 -2x ----->(1)

as this point (x,y) (the tangent point ) is also in the curve , so

y= -x^3 +3x^2 -2x ----------->(2)
so (1) & (2) are equal to y
y= -x^3 +3x^2 -2x = -3x^3+6x^2 -2x
2x^3 -3x^2 =0
x^2 (2x-3) =0
x=0
or
x=3/2
when x=0 m= -2 ------->not +ve
when x=3/2 m=-3 * (3/2)^2 +6(3/2)-2 = 0.25 ---->+ve

so there is only 1 value for m that make the line tangent to the curve with +ve m
so the line equation is
y = 0.25 x

Answer 2

My calculus is admittedly rusty but here's what my creaking mind has come up with...

y= x(x-1)(2-x) so...
y= -x^3 + 3x^2 -2x so...
dy/dx = -3x^2 + 6x - 2 so...
The slope of y= x(x-1)(2-x) at the point x=0 is -2.

That means that the only straight line that's tangent to y at the point x=0 is the line with equation y = -2x + b. Since we're told that the straight line in question goes through the origin (x=0, y=0), we know that b must be 0. So, the straight line has equation y= -2x and m=-2. So the answer is that there are no straight lines with positive slope m that are tangent to the curve at the origin.