Question

The slope of normal to the curve \(y = 3x^2-6x\) at x = 0 is:

• A. -6
• B. \(-\frac{1}{6}\)
• C. \(\frac{1}{6}\)
• D. \(-\frac{2}{3}\)

Answer 1

The Correct Option is C

To find the slope of the normal to the curve \(y=3x^2-6x\) at \(x=0\), we follow these steps:

1. Determine the derivative of \(y\): 
The derivative \(y'\) gives the slope of the tangent line at any point.

\(y=3x^2-6x\)

\(y'=\frac{d}{dx}(3x^2-6x)=6x-6\)

2. Find the slope of the tangent at \(x=0\): 
Substitute \(x=0\) into the derivative.

\(y'(0)=6(0)-6=-6\)

3. Calculate the slope of the normal: 
The slope \(m\) of the normal line is the negative reciprocal of the slope of the tangent line.

If the slope of the tangent is \(-6\), then

\(m=-\frac{1}{-6}=\frac{1}{6}\)

Thus, the slope of the normal to the curve at \(x=0\) is \(\frac{1}{6}\).