A curve is defined by the equation y = x^2 - 5x + 6. What is the x-intercept of the line n… — JAMB UTME

A curve is defined by the equation $y = x^2 - 5x + 6$ . What is the $x$ -intercept of the line normal to the curve at the point where $x = 1$ ?

Cevap: -5

The x-intercept of the normal line to the curve at x = 1 is -5.

Substituting

x = 1

into

y = x^2 - 5x + 6

yields

y = 2

, identifying the point

(1, 2)

. Differentiating gives

\frac{dy}{dx} = 2x - 5

, which equals

-3

x = 1

. The normal line gradient is the negative reciprocal,

\frac{1}{3}

. The line equation

y - 2 = \frac{1}{3}(x - 1)

simplifies to

x - 3y + 5 = 0

. Setting

y = 0

gives

x = -5

Calculate the y-coordinate at x = 1 to determine the point of tangency

Substituting

x = 1

into

y = x^2 - 5x + 6

gives

y = (1)^2 - 5(1) + 6 = 2

, yielding the point

(1, 2)

The normal line intersects the curve at the point of tangency.

Find the derivative of the curve and evaluate the tangent slope

\frac{dy}{dx} = 2x - 5

. At

x = 1

m_t = 2(1) - 5 = -3

The derivative evaluated at a specific point gives the slope of the tangent line to the curve.

Compute the slope of the normal line

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

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

Construct the normal line equation and determine its x-intercept

Using point-slope form:

y - 2 = \frac{1}{3}(x - 1) \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0

. Setting

y = 0

gives

x + 5 = 0 \implies x = -5

The

x

-intercept occurs where the line crosses the

x

-axis, meaning

y = 0

The slope of the normal line to a curve

y = f(x)

x = a

is the negative reciprocal of the derivative at that point,

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

Soru