Find the y-intercept of the tangent line to the curve y = x^3 - 6x + 8 at the point where … — JAMB UTME

Find the $y$ -intercept of the tangent line to the curve $y = x^3 - 6x + 8$ at the point where $x = 1$ .

Cevap: 6

The

y

-intercept of the tangent line is 6.

Differentiating

y = x^3 - 6x + 8

yields

\frac{dy}{dx} = 3x^2 - 6

. At

x = 1

, the point on the curve is

(1, 3)

and the gradient of the tangent is

m = -3

. Substituting into the point-slope form

y - 3 = -3(x - 1)

gives

y = -3x + 6

. Setting

x = 0

identifies the

y

-intercept as 6.

Find the point of tangency on the curve

For

x = 1

y = (1)^3 - 6(1) + 8 = 3

. The point is

(1, 3)

The tangent line touches the curve at the point corresponding to

x = 1

Find the gradient function of the curve

\frac{dy}{dx} = 3x^2 - 6

The derivative of the curve equation gives the gradient of the tangent at any point.

Calculate the gradient of the tangent line at

x = 1

m = 3(1)^2 - 6 = -3

Substitute

x = 1

into the derivative.

Formulate the equation of the tangent line

y - 3 = -3(x - 1) \implies y = -3x + 6

Apply the point-slope line equation

y - y_1 = m(x - x_1)

using

(1, 3)

and

m = -3

Determine the

y

-intercept

Setting

x = 0

yields

y = 6

The

y

-intercept is the value of

y

when

x = 0

Tangents to Curves and Axis Intercepts

Soru