Question

The slope of the normal to the curve \( y = x^3 - 3x^2 + 2x + 1 \) at the point where the tangent is horizontal is

• A. \( 0 \)
• B. \( 1 \)
• C. \( -1 \)
• D. \( \infty \)
• E. \( -\infty \)

Hint:

Horizontal tangent ⇒ slope = 0 ⇒ normal is vertical ⇒ slope = \( \infty \).

Answer 1

The Correct Option is D

Concept:
• A tangent is horizontal when: \[ \frac{dy}{dx} = 0 \]
• The slope of the normal is the negative reciprocal of the slope of the tangent: \[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} \]
• If slope of tangent is \(0\), then: \[ m_{\text{normal}} = \infty \]

Step 1: Write the given function.
\[ y = x^3 - 3x^2 + 2x + 1 \]

Step 2: Differentiate the function.
Differentiate term-by-term: \[ \frac{d}{dx}(x^3) = 3x^2 \] \[ \frac{d}{dx}(-3x^2) = -6x \] \[ \frac{d}{dx}(2x) = 2 \] \[ \frac{d}{dx}(1) = 0 \] Thus: \[ \frac{dy}{dx} = 3x^2 - 6x + 2 \]

Step 3: Find where tangent is horizontal.
Horizontal tangent condition: \[ \frac{dy}{dx} = 0 \] So: \[ 3x^2 - 6x + 2 = 0 \]

Step 4: Solve quadratic equation.
\[ 3x^2 - 6x + 2 = 0 \] Use quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(2)}}{2\cdot3} \] \[ = \frac{6 \pm \sqrt{36 - 24}}{6} \] \[ = \frac{6 \pm \sqrt{12}}{6} \] \[ = \frac{6 \pm 2\sqrt{3}}{6} \] \[ = 1 \pm \frac{\sqrt{3}}{3} \]

Step 5: Interpret the result.
At both these values of \(x\):
• Tangent slope = 0
• Tangent line is horizontal

Step 6: Find slope of normal.
Using relation: \[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} \] Since: \[ m_{\text{tangent}} = 0 \] Thus: \[ m_{\text{normal}} = -\frac{1}{0} \] \[ = \infty \]

Step 7: Geometric meaning.

• Horizontal tangent ⇒ flat line
• Normal is perpendicular ⇒ vertical line
• Vertical line has infinite slope

Step 8: Final Answer.
\[ \boxed{\infty} \]