# `f(x) = 2x^2, 4x + y + 3 = 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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### 2 Answers

Since the tangent line to function f(x) = 2x ^2 has to be parallel to the line 4x + y + 3 = 0, both needs to have the same slope.

Equation of the straight line can be rewritten in slope-intercept form to : y = -4x - 3. Hence the desired slope of our line is -4.

And derivative of f(x) will provide the slope of the line tangent to f(x)

d/dx(f(x)) = 4x

setting it equal to -4

4x = -4 or x= -1 provides one of the conditions

Corresponding y coordinate can be found from the f(x) = 2x^2 equation, y = 2*(-1)^2 = 2

The slope of the desired line is equal to -4, and we now have a point on the desired line (-1,2), as well the point of tangency

Using the point-slope, equation to desired line can be obtained as follows. Given slope m = -4 and a point on the desired line,(x0,y0), the equation of the line is given by

(y - y0) = m(x - x0)

y -2 = -4(x + 1 )

y = -4x -2

The given line is :-

4x + y + 3 = 0

or, y = -4x - 3 (the line is represented in slope intercept form)

Thus, the slope of the line = -4

Now, the tangent to the curve f(x) = 2(x^2) is parallel to the above line

Thus, the slope of the tangent = slope of the line = -4.......(1)

The given function is:-

f(x) = 2(x^2)

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

f'(x) = 4x

Now, slope of the tangent = -4

Thus, 4x = -4

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

f(-1) = y = 2

Hence the tangent passes through the point (-1,2)

Thus, equation of the tangent at the point (-1,2) and having slope = -4 is :-

y - 2 = (-4)*(x - (-1))

or, y - 2 = -4x - 4

or, y + 4x + 2 = 0 is the equation of the tangent to the given curve at (-1,2)