Question

Find the equations of the tangent and normal to the given curves at the indicated points: 

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)

(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3)

Answer 1

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy 

\(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx 

slope at (0,5) = -10 

∴ Equation of tangent at (0, 5) is 

y – 5 = -10 (x – 0) 

y – 5 = -10 x 10 x + y – 5 = 0 

slope of the normal at (0,5)

(0, 5) = \(\frac{-1}{-10} = \frac{1}{10}\) 

∴ equation of normal is 

y – 5 = \(\frac{1}{10}\)(x – 0) = 10y – 50 = x 

x – 10y + 50 = 0

(ii) \(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx 

slope of the tangent at x = 1