What is the gradient of the normal to the curve y = x^2 + 2x - 1 at the point where x = 1? — JAMB UTME

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

The gradient of the normal to the curve at

x = 1

-\frac{1}{4}

Differentiating

y = x^2 + 2x - 1

gives

\frac{dy}{dx} = 2x + 2

. Substituting

x = 1

yields a tangent slope of

4

. Taking the negative reciprocal gives

-\frac{1}{4}

, which is the correct gradient of the normal.

Find the derivative of the curve equation to determine the general tangent gradient function.

\frac{dy}{dx} = 2x + 2

The derivative of a function gives the slope of the tangent line at any point

x

Evaluate the tangent gradient at the specific point

x = 1

m = 2(1) + 2 = 4

Substituting

x = 1

into the derivative gives the slope of the tangent at that specific x-coordinate.

Calculate the gradient of the normal line using the perpendicular condition.

m_n = -\frac{1}{m} = -\frac{1}{4}

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

Relationship between tangent and normal gradients (

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}

)

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