What is the y-intercept of the normal line to the curve y = x^3 - 3x^2 + 4x - 1 at the poi… — JAMB UTME

What is the $y$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ ?

The

y

-intercept of the normal line is

2

Evaluating

y = x^3 - 3x^2 + 4x - 1

x = 1

yields

y = 1

. Differentiating gives

\frac{dy}{dx} = 3x^2 - 6x + 4

, which evaluates to

1

x = 1

. Since the normal is perpendicular to the tangent, its gradient is

-1

. Substituting into

y - 1 = -1(x - 1)

yields

y = -x + 2

, giving a

y

-intercept of

2

Find the

y

-coordinate of the point of tangency.

x = 1

y = (1)^3 - 3(1)^2 + 4(1) - 1 = 1 - 3 + 4 - 1 = 1

. Point of contact is

(1, 1)

The point must lie on the curve.

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 3x^2 - 6x + 4.

The first derivative represents the gradient of the tangent to the curve.

Calculate the gradient of the tangent and normal at

x = 1

Tangent gradient

m_t = 3(1)^2 - 6(1) + 4 = 1

. Normal gradient

m_n = -\frac{1}{m_t} = -1

The normal line is perpendicular to the tangent line, so

m_n \cdot m_t = -1

Determine the equation of the normal line and find its

y

-intercept.

Using

y - y_1 = m_n(x - x_1) \implies y - 1 = -1(x - 1) \implies y = -x + 2

. Setting

x = 0

gives

y = 2

The

y

-intercept occurs where the line crosses the

y

-axis (

x = 0

Tangents and Normals to Curves

Question