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CALCULUS HELP PLEASE!!!!?

+1 vote

What is the equation of a line tangent to the graph of y= x^(3) + 3x^(2) + 2 at its point of inflection?

Please help I'm having so much trouble

asked Jan 28, 2013 in CALCULUS by dkinz Apprentice

1 Answer

+2 votes

y = x3 + 3x2 + 2

The point of inflection occurs when y'' = 0.

Apply derivative with respect to x each side.

(d/dx)y = (d/dx)(x3 + 3x2 + 2)

y' = (d/dx)(x3) + (d/dx)(3x2) +(d/dx)( 2)

Recall Power Rule for Function is (d/dx)(xn) = n xn-1

General Derivative Formula is (dc/dx) = 0 where c is constant

y' = 3x2 + 3(2x) + 0

y' = 3x2 + 6x

Again apply derivative with respect to x each side.

y'' = 6x + 6

If y'' = 0 then 6x + 6 = 0

So x + 1 = 0

Subtract 1 from each side.

x = -1

When x = -1, y = (-1)3 + 3(-1)2 + 2 = -1 + 3 + 2 = 4

 So the inflection point is (-1, 4).

 

At x = -1, the slope of the tangent line is y' = 3(-1)2 + 6(-1) = 3 - 6 = -3.

Recall point slope form is y - y1 = m (x - x1)

Point(x1, y1) = (-1, 4) and slope m = - 3

So the equation of the tangent line at the point of inflection is:

y - 4 = -3(x + 1)

y - 4 = - 3x - 3

Add 4 to each side

y = - 3x + 1

The equation of the tangent line at the point of inflection is y = - 3x + 1

answered Jan 28, 2013 by richardson Scholar

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