What is the y-intercept of the tangent line to the curve y = x^3 - 2x + 4 at the point whe… — JAMB UTME

What is the $y$ -intercept of the tangent line to the curve $y = x^3 - 2x + 4$ at the point where $x = 1$ ?

Answer: 2

The y-intercept of the tangent line is 2.

The curve evaluated at

x=1

gives point

(1,3)

. The derivative

y'=3x^2-2

gives slope

m=1

x=1

. The tangent line equation is

y-3=1(x-1)

, which simplifies to

y=x+2

. The

y

-intercept occurs at

x=0

, giving

y=2

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

x = 1

x = 1

y = (1)^3 - 2(1) + 4 = 3

. The point of tangency is

(1, 3)

The point of tangency lies on the curve.

Find the gradient function of the curve by differentiation.

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

The first derivative represents the slope of the tangent line.

Calculate the slope of the tangent line at

x = 1

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

Substituting

x = 1

into the derivative yields the slope at that specific point.

Formulate the equation of the tangent line.

y - 3 = 1(x - 1) \implies y = x + 2

Use point-slope form

y - y_1 = m(x - x_1)

with point

(1,3)

and slope

m=1

Find the

y

-intercept of the tangent line.

Setting

x = 0

gives

y = 2

The

y

-intercept is the value of

y

where the line crosses the vertical axis.

Tangents and Normals to Curves

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