Find an equation for the line perpendicular to the tangent line to the curve y = x³ - 4x + 1 at the point (2, 1).

Solution:

The slope of the of the tangent line to the given curve y = x³ - 4x + 1 at the point (2,1) can be found out by differentiating 

dy/dx = 3x² - 4

Slope of the tangent at (2, 1) = 3(2)² - 4 = 8

Beginning with the slope of the secant passing through points P(xo, f(xo)) and Q(xo+h, f(xo+h).

Point Q is not shown in the figure below as it is close to point P

Hence the slope of the line L perpendicular to the tangent is -1/8.

Equation of the perpendicular line passing through point (2,1)

y - 1 = m(x - 2)

Substituting m = -1/8 in the equation above we get

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

8y - 8 = -x + 2

y = (1/8)(10 - x)

y = -x/8 + 5/4

Find an equation for the line perpendicular to the tangent line to the curve y = x³ - 4x + 1 at the point (2, 1).

Summary:

The equation for the line perpendicular to the tangent line to the curve y = x³ - 4x + 1 at the point (2, 1) is y = -x/8 + 5/4