Calculus of a Single Variable, Chapter 2, 2.1, Section 2.1, Problem 33

The given line is :-
2x - y + 1 = 0
or, y = 2x + 1 (the line is represented in slope intercept form)
Thus, the slope of the line = 2
Now, the tangent to the curve f(x) = (x^2) is parallel to the above line
Thus, the slope of the tangent = slope of the line = 2.......(1)
The given function is:-
f(x) = (x^2)
differentiating both sides w.r.t 'x' we get
f'(x) = 2x
Now, slope of the tangent = 2
Thus, 2x = 2
or, x = 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 (2,4) and having slope = 2 is :-
y - 1 = (2)*(x - 1)
or, y - 1 = 2x - 2
or, y - 2x + 1 = 0 is the equation of the tangent to the given curve at (1,1)