Calculus of a Single Variable, Chapter 2, 2.1, Section 2.1, Problem 34

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)