SUMMARY
The discussion focuses on finding the equation of a line perpendicular to the tangent of the curve defined by the equation y=x^3-4x+1 at the point (2,1). The derivative of the curve is calculated as y' = 3x^2 - 4, yielding a slope of 8 at x=2. The tangent line is determined to be y=8x-15. The correct perpendicular line, which must pass through the point (2,1), is derived as y=-x/8 + 5/4, correcting the initial miscalculation of the y-intercept.
PREREQUISITES
- Understanding of calculus, specifically derivatives and slopes of curves
- Familiarity with the point-slope form of a linear equation
- Knowledge of perpendicular lines and their slopes
- Basic algebra skills for manipulating equations
NEXT STEPS
- Study the concept of derivatives in calculus, focusing on applications in finding tangents
- Learn about the point-slope form of linear equations and how to apply it
- Explore the relationship between slopes of perpendicular lines
- Practice solving problems involving tangent and normal lines to curves
USEFUL FOR
Students studying calculus, particularly those focusing on derivatives and their applications in geometry, as well as educators looking for examples of tangent and perpendicular lines.
