What is the equation of the normal line to the curve y = x^3 - 3x + 2 at the point where x… — JAMB UTME

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

The equation of the normal line is

x + 9y - 38 = 0

Evaluating

y

x = 2

gives the point

(2, 4)

. Differentiating

y = x^3 - 3x + 2

yields

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

. At

x = 2

, the tangent slope is

m_t = 9

, making the normal slope

m_n = -1/9

. Using the point-slope formula

y - 4 = -1/9(x - 2)

simplifies to

x + 9y - 38 = 0

Find the

y

-coordinate at

x = 2

y = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4

. The point on the curve is

(2, 4)

The point of tangency/normal line intersection must be determined on the curve.

Calculate the derivative

\frac{dy}{dx}

to find the slope of the tangent line

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

. At

x = 2

\frac{dy}{dx} = 3(2)^2 - 3 = 12 - 3 = 9

The derivative evaluated at

x = 2

gives the gradient of the tangent line.

Find the slope of the normal line

m_n = -\frac{1}{m_t} = -\frac{1}{9}

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

Use the point-slope form to find the equation of the normal line

y - 4 = -\frac{1}{9}(x - 2) \implies 9(y - 4) = -(x - 2) \implies 9y - 36 = -x + 2 \implies x + 9y - 38 = 0

Substituting point

(2, 4)

and slope

-1/9

into

y - y_1 = m_n(x - x_1)

yields the standard line equation.

The slope of the normal line to a curve

y = f(x)

(x_1, y_1)

m_n = -\frac{1}{f'(x_1)}

Question