Question

Find the equation of the tangent and the normal to the following curves at the indicated points: 

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

Answer 1

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=4x^3-18x^2+26x-10\)

m(tangent) at (0,5) = – 10

m(normal) at (0, 5) = \(\frac{1}{10}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 5 = – 10x

y + 10x = 5

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-5=\frac{1}{10}x\)