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

What is the gradient of the normal line to the curve $y = x^2 - 3x + 5$ at the point where $x = 1$ ?

Answer: 1

The gradient of the normal line to the curve at

x = 1

is 1.

Differentiating

y = x^2 - 3x + 5

gives

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

. Evaluating this derivative at

x = 1

gives the tangent gradient

m_t = -1

. Because the normal line is perpendicular to the tangent, its gradient is

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

Differentiate the function with respect to x

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

The derivative of a function gives the slope of the tangent line at any given x-coordinate.

Evaluate the derivative at x = 1

m_t = -1

Substituting the given point's x-coordinate into the gradient function yields the slope of the tangent.

Calculate the negative reciprocal of the tangent slope

m_n = 1

Since the normal line is perpendicular to the tangent line, its gradient is m_n = -1 / m_t.

Gradient of a Normal Line

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