Question

question 2 of 20
you may not use your calculator for this question.
let ( f(x) = x^3 - 4 ). which of these is the equation for the normal line to this curve at the point ( (2, 4) )?

a. ( y = -\frac{1}{12}x + \frac{25}{6} )

b. ( y = 12x - \frac{25}{6} )

c. ( y = \frac{1}{12}x - \frac{25}{6} )

d. ( y = -\frac{1}{12}x - 2 )

e. ( y = -12x - 4 )

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = x^3 - 4 \). The derivative \( f'(x) \) using the power rule (\( \frac{d}{dx}x^n = nx^{n - 1} \)) is \( f'(x)=3x^2 \).

Step2: Evaluate the derivative at \( x = 2 \)

Substitute \( x = 2 \) into \( f'(x) \): \( f'(2)=3\times(2)^2 = 3\times4 = 12 \). This is the slope of the tangent line at \( (2,4) \).

Step3: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, if the tangent slope is \( 12 \), the normal slope \( m \) is \( -\frac{1}{12} \).

Step4: Use point - slope form to find the normal line equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(2,4) \) and \( m = -\frac{1}{12} \).
Substitute the values: \( y - 4=-\frac{1}{12}(x - 2) \).
Expand the right - hand side: \( y - 4=-\frac{1}{12}x+\frac{2}{12}=-\frac{1}{12}x+\frac{1}{6} \).
Add \( 4 \) to both sides: \( y=-\frac{1}{12}x+\frac{1}{6}+4 \).
Since \( 4=\frac{24}{6} \), then \( y = -\frac{1}{12}x+\frac{1 + 24}{6}=-\frac{1}{12}x+\frac{25}{6} \).

Answer:

A. \( y = -\frac{1}{12}x+\frac{25}{6} \)