# `f(x) = x^3, 3x - y + 1 = 0` Find an equation of the line that is tangent to the graph of f and parallel to the given line.

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Since the tangent line to function f(x) = x ^3 has to be parallel to the line 3x - y + 1 = 0, both needs to have the same slope.

Equation of the straight line can be rewritten in slope-intercept form to : y = 3x - 1. 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 equation, y = (1)^3 = 1

The slope of the desired line is equal to 3, and we now have a point on the desired line (1,1), 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 - 1 = 3(x - 1 )

y = 3x -2

The given line is :-

3x - y + 1 = 0

or, y = 3x + 1 (the line is represented in slope intercept form)

Thus, the slope of the line = 3

Now, the tangent to the curve f(x) = (x^3) is parallel to the above line

Thus, the slope of the tangent = slope of the line = 3.......(1)

The given function is:-

f(x) = (x^2)

differentiating both sides w.r.t 'x' we get

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

Now, slope of the tangent = 3

Thus, 3(x^2) = 3

or, x = +1 or -1

Putting the value of x =1 in the given equation of curve, we get

f(1) = y = 1

Hence the tangent passes through the point (1,1)

Thus, equation of the tangent at the point (1,1) and having slope = 3 is :-

y - 1 = (3)*(x - 1)

or, y - 1 = 3x - 3

or, y - 3x + 2 = 0 is the equation of the tangent to the given curve at (1,1).........(2)

**Putting x = -1 in the equation of curve we get**

**f(-1) = y = -1**

**Thus, the equation of the tangent passing through the point (-1,-1) and having slope 3 will be:-**

**y - (-1) = 3(x - (-1))**

**or, y + 1 = 3x + 3**

**or, y - 3x - 2 = 0**