Question

Tangents to the curve y = x³ at x = -1 and x = 1 are:

Answer 1

the curve is

y = x^3

the tangent on this curve is given by

m = dy/dx = d x^3/dx = 3x^2

when x = -1

dy/dx = 3*(-1)^2 = 3

moreover, when x= -1, y = (-1)^3 = -1

the equation of tangent is given by

y = mx + c

y = 3x + c

this slope touches the point (-1,-1)

-1 = 3*(-1) + c

-1 = -3 + c

c = -1 + 3

c = 2

the equation of slope at x = -1 is

y = 3x + 2

when x = 1

y = x^3 = 1^3 = 1

m = dy/dx = 3*1^2 = 3

equation of slope

y = mx + c

y = 3x + c

this slope touches the point (1,1)

1 = 3*1 + c

1 = 3 + c

c = 1 - 3

c = -2

equation of slope at x = 1 is

y = 3x - 2

Answer 2

Answer:

Step-by-step explanation: