The first derivative of a curve at any point gives the slope of the tangent at that point.

Here the curve is defined by f(x) = x^2 - 2x + 2

f'(x) = 2x - 2

As the tangent is perpendicular to the the line 4y - x - 5...

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The first derivative of a curve at any point gives the slope of the tangent at that point.

Here the curve is defined by f(x) = x^2 - 2x + 2

f'(x) = 2x - 2

As the tangent is perpendicular to the the line 4y - x - 5 = 0, it has a slope that is the negative reciprocal of the slope of the line.

4y - x - 5 = 0

=> y = x/4 + 5/4

The slope of the line is 1/4. The slope of the tangent is -4

So 2x - 2 = -4

=> 2x = -2

=> x = -1

For x = -1, f(x) = (-1)^2 + 2 + 2 = 5

Therefore the coordinates of the tangential point are **(-1,5)**