SOLUTION: Find the equation of the line that passes through point (0,5) and is tangent to the curve {{{ y=x^3 - 2x^2 + 1}}}.

Let the point of tangency be (p,q)

Then since it must satisfy y=x³-2x²+1 

*  The point of tangency is (p,p³-2p²+1)

Let the equation of the tangent line be y = mx+b

Since the tangent line goes through (0,5),

    5 = m(0)+b
    5 = 0+b
    5 = b

*  Equation of tangent line is y = mx+5

The slope of the tangent line is the derivative of
y=x³-2x²+1 at the point of tangency:
y'=3x²-4x 

we substitute y'=m and x=p

m = 3p²-4p

So the equation of the tangent line y = mx+5 is now

*    y = (3p²-4p)x+5

Since the tangent line must go through the point
of tangency (p,p³-2p²+1)

p³-2p²+1 = (3p²-4p)p+5

p³-2p²+1 = 3p³-4p²+5
       0 = 2p³-2p²+4
       0 = p³-p²+2
     
p = -1 is the only real solution.

We find the equation of the tangent line with p = -1

y = (3p²-4p)x+5

y = [3(-1)²-4(-1)]x+5

y = [3(1)+4]x+5

y = [3+4]x+5

y = 7x+5


   

Edwin