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)