Find the slope of the tangent line to the parabola y = 4x - x² at the point (1, 3)

Solution:

Given, equation of parabola y = 4x - x²

Point (1, 3)

The tangent line to a curve is the one that coincides with the curve at a point and with the same derivative, i.e. the same degree of variation.

Differentiating the equation of parabola,

y’ = 4 - 2x

Substituting the point (1, 3) in the above equation,

y(1)’ = 4 - 2(1)

y(1)’ = 4 - 2 = 2

The equation of the line is given by 

y - f(a) = f’(a) (x - a)

Now, y - 3 = 2 (x - 1)

y - 3 = 2x - 2

y = 2x - 2 + 3

y = 2x + 1

The equation of the tangent line is y = 2x + 1.

The above equation represents the slope-intercept form of a line y = mx + c

So, slope m = 2

Therefore, the slope of the tangent line is 2.

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Find the slope of the tangent line to the parabola y = 4x - x² at the point (1, 3)

Summary:

The slope of the tangent line to the parabola y = 4x - x² at the point (1, 3) is 2.