Please help?

Question

Find the Tangents
y=x^3-2x^2+x+5
find the equation for the tangents when y=5

Answer 1

if y=5, then... x^3-2x^2+x=0 or x(x^2-2x+1) = 0

roots are x = 0 , x^2-2x+1 = 0.....(x-1)^2= 0...x = 1

to find Eq of the line use : Y-Y1= m(X-X1)

Answer 2

Take the derivative of the function. That's teh equation for the slope of a tangent to the curve at any point (x, f(x)) on it.

By inspection, the given function is 5 when x^3-2x^2+x=0 so the roots are x=0 and x=1. Plug those values into the derivative to get the slope at that point and then fit it through the points (0, 5) and (1, 5) and you're done.

HTH

Doug

Answer 3

When x = 5, y = 5^3 - 2*5^2 + 5 + 5 = 85. The derivative is y' = 3x^2 - 4x + 1, so when x = 5, y' = 36. The tangent is the line through (5,85) with slope 36.