Question

Find the equation of the tangent and the normal to the given curve at the indicated point: y = x³ – 2x + 7 at (1, 6)

Answer 1

Consider y = x³ – 2x + 7 as the equation of curve

By differentiating both sides w.r.t x

dy/dx = 3x² - 2

We know that

Where m = 1

So the required equation of the tangent at point (1, 6),

(y – y₁) = m(x – x₁)

By further simplification

(y – y₁)/(x – x₁) = m

By substituting the values

(y – 6)/(x – 1) = 1

By cross multiplication

y – 6 = x – 1

So we get

y – 6 + 1 = x

y – 5 = x

It can be written as

x – y + 5 = 0

Here the required equation of normal at point (1, 6)

(y – y₁) = -1/m(x – x₁)

By cross multiplication

(y – y₁)/(x – x₁) = -1/m

Substituting the values

(y – 6)/(x – 1) = -1/1

It can be written as

y – 6 = -x + 1

By further simplification

x + 2y – 6 – 1 = 0

x + y – 7 = 0