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Question

This problem is killing me:

y = x^3 - 3x^2 + 4 P(2,0)

Find the slope of the curve at P and the equation of the tangent line at P.

Any help PLEASE, i'm dying from this...............................................................

Answer 1

Take the derivative:

y' = 3x² - 6x

Evaluate it at x = 2:

y'(2) = 3(2)² - 6(2) = 0

So the slope of the tangent line will be 0:

y - 0 = 0(x - 2)

Which results in the horizontal line:

y = 0

Answer 2

To find slope of a curve find it's derivative. In this case U need to know that the derivative of the "x ^ n" is "n * (x ^ n-1)". So the derivative of the y = x^3 - 3x^2 + 4 is
dy/dx = 3 * (x ^ 2) - 6 * x

slope of the curve can be found by substituting x=2 to the above equation. (because x coordiante of the p is 2)

slope = 0

equation of the tangent line at P is y=0 because slope is 0.

Answer 3

The slope is the value of the derivative at the point
y´= 3x^2 -6x = 3*4-6*2 =0
the tangent is horizontal
eq. y=0

Answer 4

dy/dx = 3x^2 - 6x

m(2,0) = 3*2^2 - 6*2 = 0

0 = 0*2 + b
b=0

y=0 is the tangent

Answer 5

take the derivative, y' = 3x^2 - 6x

plug in P(2,0)

Answer: 0

(the slope is horizontal at this point)

Answer 6

dy/dx = (3x^2 - 6x)
therefore slope at (2,0) =dy/dx at (2,0)
slope = 0
it means equation of x axis i.e. y = 0