What is the tangent of y=4x-x^2,(1,3) ?

The tangent of y=4x-x^2,(1,3) is y=2x+1

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tangent of y=4x-x^2,(1,3)

tangent of

y = 4 x - x^{2}, (1, 3)

y = 2 x + 1

Solution steps

Find the slope of

y = 4 x - x^{2} : \frac{d y}{d x} = 4 - 2 x

Compute the slope of

y = 4 x - x^{2}

(1, 3) : m = 2

Find the line with slope m=

2

and passing through

(1, 3) : y = 2 x + 1

y = 2 x + 1

\frac{d y}{d x} + 8 y = 10

f (x) = 3 x - 4 x^{2}, (- 3, - 45)

\frac{d}{d x} ((x - 3)^{3} (2 x - 1)^{2})

f^{^{'}} (x) = sin (4) (4 x^{5})