calculus question

Question

Find an equation of a line that is tangent to the graph of f and parallel to the given line.

Function		Line
f(x) = x3		12x − y + 6 = 0
y=	(smaller y-intercept)
y=	(larger y-intercept)

Answer 1

The given line has a slope of 12 (y = 12x + 6).

That means we want to determine when f'(x) = 12, since the 1st derivative is a formula for the slope of tangent lines to a given function.

f'(x) = 3x²

Solve 3x² = 12

x² = 4

x = ±2

When x = 2:

f(2) = 2³ = 8

So (2,8) is the point of tangency with a slope of 12.

y - 8 = 12(x - 2)

y = 12x - 16

When x = -2:

f(-2) = (-2)³ = -8

(-2, -8) is the point of tangency with a slope of 12:

y + 8 = 12(x + 2)

y = 12x +16

This graph shows those results:

desmos.com/calculator/n0ssbf4mqo