What is the gradient of the normal to the curve y = x^2 - 4x + 5 at the point where x = 3? — JAMB UTME

What is the gradient of the normal to the curve $y = x^2 - 4x + 5$ at the point where $x = 3$ ?

The gradient of the normal to the curve is

-\frac{1}{2}

Differentiating

y = x^2 - 4x + 5

gives

\frac{dy}{dx} = 2x - 4

. Substituting

x = 3

yields a tangent gradient of

2

. Since the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal,

-\frac{1}{2}

Differentiate the equation of the curve to determine the gradient function.

\frac{dy}{dx} = 2x - 4

The first derivative represents the gradient of the tangent to the curve at any point

x

Evaluate the derivative at

x = 3

to find the tangent gradient

m_t

m_t = 2(3) - 4 = 2

Substituting

x = 3

gives the slope of the tangent line at the given point.

Calculate the gradient of the normal

m_n

using

m_n = -\frac{1}{m_t}

m_n = -\frac{1}{2}

The normal line is perpendicular to the tangent line, so its gradient is the negative reciprocal of the tangent's gradient.

Gradient of a Normal Line

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