A curve is defined by the equation y = x^3 - 4x + 3. What is the y-intercept of the tangen… — JAMB UTME

A curve is defined by the equation $y = x^3 - 4x + 3$ . What is the $y$ -intercept of the tangent line to the curve at the point where $x = 2$ ?

Answer: -13

The y-intercept of the tangent line is

-13

x = 2

, substituting into

y = x^3 - 4x + 3

yields

y = 3

. The derivative

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

evaluated at

x = 2

gives

m = 8

. The tangent line equation is

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

. Setting

x = 0

gives the

y

-intercept of

-13

Calculate the y-coordinate of the point on the curve at

x = 2

y = 3

, giving the point

(2, 3)

The point of contact must lie on the curve.

Differentiate the curve equation to find the gradient function

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

The derivative of a function gives the gradient of the tangent at any point

x

Evaluate the derivative at

x = 2

to find the gradient of the tangent

m

m = 8

Substituting

x = 2

into

3x^2 - 4

yields

3(4) - 4 = 8

Formulate the linear equation of the tangent line using point

(2, 3)

and gradient

m = 8

y = 8x - 13

Applying

y - y_1 = m(x - x_1)

gives

y - 3 = 8(x - 2)

, which simplifies to

y = 8x - 13

Extract the y-intercept of the tangent line

y-intercept =

-13

In slope-intercept form

y = mx + c

, the constant term

c = -13

is the y-intercept.

Tangents to Curves and Line Intercepts

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