Since the tangent line to function f(x) = x ^3 + 2 has to be parallel to the line 3x - y - 4 = 0, both needs to have the same slope.
Equation of the straight line can be rewritten in slope-intercept form to : y = 3x - 4. Hence the desired slope of our line is 3.
And derivative of f(x) will provide the slope of the line tangent to f(x)
d/dx(f(x)) = 3x
setting it equal to 3
3x = 3 or x = 1 provides one of the conditions
Corresponding y coordinate can be found from the f(x) = x^3 + 2 equation, y = (1)^3 + 2 = 3
The slope of the desired line is equal to 3, and we now have a point on the desired line (1, 3), as well the point of tangency
Using the point-slope, equation to desired line can be obtained as follows. Given slope m = 3 and a point on the desired line,(x0, y0), the equation of the line is given by
(y - y0) = m(x - x0)
y - 3 = 3(x - 1 )
y = 3x