What is the x-intercept of the normal line to the curve y = x^2 - 4x + 5 at the point wher… — JAMB UTME

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

The

x

-intercept of the normal line is 7.

Differentiating

y = x^2 - 4x + 5

gives

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

. Substituting

x = 3

yields a tangent gradient of

2

, so the normal gradient is

-\frac{1}{2}

. The curve passes through

(3, 2)

x = 3

. The equation of the normal is

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

, which simplifies to

x + 2y - 7 = 0

. Setting

y = 0

gives

x = 7

Find the

y

-coordinate corresponding to

x = 3

y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2

. The point on the curve is

(3, 2)

The normal line passes through the specific point of tangency on the curve.

Calculate the gradient of the tangent and normal lines at

x = 3

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

. At

x = 3

m_t = 2(3) - 4 = 2

. Therefore,

m_n = -\frac{1}{m_t} = -\frac{1}{2}

The normal line is perpendicular to the tangent line.

Determine the equation of the normal line

y - 2 = -\frac{1}{2}(x - 3) \implies 2(y - 2) = -(x - 3) \implies x + 2y - 7 = 0

Use the point-slope form

y - y_1 = m(x - x_1)

to write the linear equation.

Find the

x

-intercept of the normal line

Set

y = 0

x + 2(0) - 7 = 0 \implies x = 7

The

x

-intercept occurs where

y = 0

Tangents and Normals to Curves

Estimated Time:1m 30s

Question