Question:medium
The slope of the normal to the curve \( y = x^3 - 3x^2 + 2x + 1 \) at the point where the tangent is horizontal is
The slope of the normal to the curve \( y = x^3 - 3x^2 + 2x + 1 \) at the point where the tangent is horizontal is
Show Hint
Horizontal tangent ⇒ slope = 0 ⇒ normal is vertical ⇒ slope = \( \infty \).
Updated On: May 8, 2026
- \( 0 \)
- \( 1 \)
- \( -1 \)
- \( \infty \)
- \( -\infty \)
Show Solution
The Correct Option is D
Solution and Explanation
Was this answer helpful?
0
Top Questions on Tangents and Normals
- The value of \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \) is equal to:
- MHT CET - 2024
- Mathematics
- Tangents and Normals
- Find the equation of the tangent to the curve $ y = x^3 - 3x + 1 $ at the point where $ x = 2 $.
- BITSAT - 2025
- Mathematics
- Tangents and Normals
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:
- JEE Main - 2024
- Mathematics
- Tangents and Normals
- A line passing through the point $ A(-2, 0) $, touches the parabola $ P: y^2 = x - 2 $ at the point $ B $ in the first quadrant. The area of the region bounded by the line $ AB $, parabola $ P $, and the x-axis is:
- JEE Main - 2025
- Mathematics
- Tangents and Normals
- Want to practice more? Try solving extra ecology questions todayView All Questions
