`f(x) = x^3 + 2, 3x - y - 4 = 0` Find an equation of the line that is tangent to the graph of f and parallel to the given line.

2 Answers

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

The slope of the line 3x-y-4=0 and the lines parallel to it will be same.




Therefore the slope(m) of the line is the coefficient of the x = 3

The slope of the tangent line to the graph f(x) is the derivative of the f(x)


Since the tangent line to the graph is parallel to the given line

Therefore 3x^2=3



The y coordinate can be found by substituting the value of x in the function.

for x=1 y=1^3+2=3

for x=-1 y=(-1)^3+2=1

The equation of the tangent line can be written using the point slope form of the equation.








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dh246 | (Level 1) Adjunct Educator

Posted on

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