Equation of the Normal Line...3

[nycmathguy]

For the function given below determine the equation of the normal line at x = 1.

f(x) = 2x^3 - 7

Find dy/dx.

dy/dx = 6x^2

Let x = 1.

6(1)^2 = 6.

The slope m = 6.

Evaluate f(x) at x = 1.

f(1) = 2(1)^3 - 7

f(1) = 2 - 7

f(1) = -5


This leads to the point of tangency (1, -5).

Find the tangent line.

y - y_1 = m(x - x_1)

y - (-5) = 6(x - 1)

y + 5 = 6x - 6

y = 6x - 6 - 5

y = 6x - 11

The slope of the normal line is the negative reciprocal of the coefficient of x.

That is, m = -1/6.

y - y_1 = m(x - x_1)

y - (-5) = (-1/6)(x - 1)

y + 5 = (-1/6)(x - 1)

y + 5 = (-1/6)(x) + 1/6

y = (-1/6)(x) + 1/6 - 5

y = (-1/6)(x) - 29/6

You say?

 

[MathLover1]

correct

 

[nycmathguy]

correct

Thank you. I hope to post a few questions today but not sure right now. I had to take my 8 hour annual security renewal course required by law. I'm a bit tired. My Saturday is almost over.