Question

1. Find the equation of the tangent to


`y = {(2x + 1)}^(4)`
at the point where x = -1.

2. Find the equation of the tangent to the curve

`y = {(2x - 1})^(8)`
at the point where x = 1.

3.Find the equation of the normal to the curve

`y = {(3x - 4})^(3)`
at (1,-1).​

Answer 1

1) y = -216(x + 1)

2) y = 16(x - 1)

3) y + 1 = (-1/12)(x - 1)

Explanation:

1) y = (2x - 1)⁴

dy/dx = 8(2x - 1)³

At x = -1, we have;

dy/dx = 8(2(-1) - 1)³

dy/dx = -216

Thus,equation is;

y = -216(x - (-1))

y = -216(x + 1)

2) Equation of tangent to a curve is simply gotten by differentiation of the curve function.

y = (2x - 1)^(8)

dy/dx = 16(2x - 1)^(7)

At x = 1;

dy/dx = 16(2(1) - 1)^(7)

dy/dx = 16

Thus, equation is;

y = 16(x - 1)

3) at normal to the curve, slope will be -1/(dy/dx)

y = (3x - 4)³

dy/dx = 12(3x - 4)²

at (1,-1);

dy/dx = 12(3(1) - 4)²

dy/dx = 12

Slope = -1/12

Equation is;

y - (-1) = -(1/12)(x - 1)

y + 1 = (-1/12)(x - 1)