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

The tangent of y=4x^2-x^3,(1,3) is y=5x-2

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

tangent of

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

y = 5 x - 2

Solution steps

Find the slope of

y = 4 x^{2} - x^{3} : \frac{d y}{d x} = 8 x - 3 x^{2}

Compute the slope of

y = 4 x^{2} - x^{3}

(1, 3) : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

(0.1) (1.1)

x \to a lim (0)

f (x) = x x + 3

\int \frac{5 x}{x} d x

\int 2 sec^{4} (\frac{x}{2}) d x

What is the tangent of y=4x^2-x^3,(1,3) ?
The tangent of y=4x^2-x^3,(1,3) is y=5x-2