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

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

Answer: 2

The

x

-intercept of the normal line is

2

x = 1

, the point on the curve

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

(1, 1)

. The derivative is

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

, which equals

1

x = 1

. The normal gradient is therefore

-1

. The normal line equation is

y - 1 = -1(x - 1)

, which simplifies to

y = -x + 2

. Setting

y = 0

gives

x = 2

Find the y-coordinate at x = 1

y = 1, so the point on the curve is (1, 1)

The point of contact is needed to construct the equation of the normal line.

Differentiate the function with respect to x

dy/dx = 3x^2 - 6x + 4

The derivative gives the expression for the gradient of the tangent to the curve.

Find the gradient of the tangent and normal at x = 1

Gradient of tangent m_t = 1; gradient of normal m_n = -1

The normal line is perpendicular to the tangent line, so m_n = -1 / m_t.

Determine the equation of the normal line

y - 1 = -1(x - 1) => y = -x + 2

Applying the straight-line equation y - y_1 = m(x - x_1).

Find the x-intercept of the normal line

x = 2

Setting y = 0 in the normal line equation yields the x-intercept.

Equation and axis intercepts of a normal line to a curve

Question