A curve has the equation y = \frac{1}{3}x^3 - 2x^2 + 3x + 1. What is the positive x-coordi… — JAMB UTME

A curve has the equation $y = \frac{1}{3}x^3 - 2x^2 + 3x + 1$ . What is the positive $x$ -coordinate of the point on the curve where the tangent line is parallel to the line $y = 8x - 5$ ?

Cevap: 5

Cevap

The positive

x

-coordinate of the point where the tangent is parallel to the line is

5

The gradient of the line

y = 8x - 5

8

. Differentiating

y = \frac{1}{3}x^3 - 2x^2 + 3x + 1

gives

\frac{dy}{dx} = x^2 - 4x + 3

. Setting

\frac{dy}{dx} = 8

leads to

x^2 - 4x - 5 = 0

, which factors as

(x - 5)(x + 1) = 0

. The solutions are

x = 5

and

x = -1

. Selecting the positive value gives

x = 5

Adım Adım Çözüm

Find the gradient (slope) of the given straight line.

The line equation is in slope-intercept form

y = mx + c

, where the slope

m = 8

Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.

Differentiate the curve equation to find the gradient function

\frac{dy}{dx}

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3

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

x

Equate the derivative to the slope of the line and solve for

x

x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0

. The roots are

x = 5

and

x = -1

Solving the quadratic equation gives all

x

-values where the tangent line has a slope of 8.

Select the positive

x

-coordinate as requested.

x = 5

The question specifically asks for the positive value among the solutions.

Anahtar Kavram

Finding points on a curve where the tangent is parallel to a given line

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