What is the equation of the normal line of f(x)=x3⋅(3x−1) at x=-2?

Question

shanna87mn · Accepted Answer

The normal line to a tangent is perpendicular to the tangent. We can find the slope of the tangent line using the derivative of the original function, then take its opposite reciprocal to find the slope of the normal line at the same point.

f (x) = 3 x^{4} - x^{3}

f^{'} (x) = 12 x^{3} - 3 x^{2}

f^{'} (- 2) = 12 {(- 2)}^{3} - 3 {(- 2)}^{2} = 12 (- 8) - 3 (4) = - 108

If -108 is the slope of the tangent line, the slope of the normal line is

\frac{1}{108}

.
The point on f(x) that the normal line will intersect is (-2, -56).
We can write the equation of the normal line in point-slope form:

y + 56 = \frac{1}{108} (x + 2)

In slope-intercept form:

y = \frac{1}{108} x - \frac{3135}{56}

What is the equation of the normal line

Answered question

Answer & Explanation

New Questions in Calculus 1