# `f(x) = x^2, 2x - 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 has to be parallel to the line 2x -y + 1 = 0, both needs to have the same slope.

Equation of the straight line can be rewritten in slope-intercept form to : y = 2x - 1. Hence the desired slope of our line is 2.

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

d/dx(f(x)) = 2x

setting it equal to 2

2x = 2 or x= 1 provides one of the conditions

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

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

Using the point-slope, equation to desired line can be obtained as follows. Given slope m = 2 and a point on the desired line,(x0,y0), the equation of the line is given by

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

y - 1 = 2(x - 1 )

y = 2x -1

Which shows that there is only one line with the point of tangency and parallel to itself for this function.

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(2) = y = 4

Hence the tangent passes through the point (2,4)

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

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

or, y - 4 = 2x - 4

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

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)

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 .Since the tangent to the curve f(x) = (x^2) is parallel to the above line the slope of the tangent = slope of the line = 2

The given function is:- f(x) = (x^2) differentiating we get f'(x) = 2x. Since the slope of the tangent = 2 Thus, 2x = 2 or, x = 1 . We use the value of x =1 in the given equation of curve and we get f(2) = y = 4 Hence the tangent passes through the point (2,4) Thus, equation of the tangent at the point (2,4) and having slope = 2 is :- y - 4 = (2)*(x - 2) or, y - 4 = 2x - 4 or, y = 2x is the equation of the tangent to the given curve at (2,4)