How do you find the equation of the line that is tangent to #f(x)=x^3+2# and parallel to the line #3x-y-4=0#?

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Answer 1 · 2016-10-13T22:22:16

Two tangent lines are parallel to the given line, #y = 3x + 4# and #y = 3x#

Write the line in slope-intercept form so that we can observe its slope:

#y = 3x - 4#

Its slope is 3.

Take the first derivative of f(x) and set that equal to 3

#f'(x) = 3x^2#

#3x^2 = 3#

#x^2 = 1#

#x = -1# and #x = 1#

The corresponding y values are:

#y = -1^3 + 2 = 1# and #y = 1^3 + 2 = 3#

The points where the tangent lines are parallel to the given line are #(-1,1)# and #(1,3)#.

Using the point-slope form of the equation of a line, we create two equations:

#y - 1 = 3(x - -1)# and #y - 3 = 3(x - 1)#

Simplify both:

#y = 3x + 4# and #y = 3x#