The slope of normal to the curve \(y = 3x^2-6x\) at x = 0 is:
- -6
- \(-\frac{1}{6}\)
- \(\frac{1}{6}\)
- \(-\frac{2}{3}\)
The Correct Option is C
Solution and Explanation
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 \(\frac{1}{6}\).
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