What is the tangent of f(x)=8x^2-x^3,(2,24) ?

The tangent of f(x)=8x^2-x^3,(2,24) is y=20x-16

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tangent of f(x)=8x^2-x^3,(2,24)

tangent of

f (x) = 8 x^{2} - x^{3}, (2, 24)

y = 20 x - 16

Solution steps

Find the slope of

f (x) = 8 x^{2} - x^{3} : \frac{df ( x )}{d x} = 16 x - 3 x^{2}

Compute the slope of

f (x) = 8 x^{2} - x^{3}

(2, 24) : m = 20

Find the line with slope m=

20

and passing through

(2, 24) : y = 20 x - 16

y = 20 x - 16

(3, 4), (5, 8)

y^{^{'}} = \frac{2 y}{3 x}

\int \frac{x + 6}{x ^{2} + 11 x + 18} d x

\frac{1}{2 + x}

t \to 0 lim ((tan (5 t)) k)

What is the tangent of f(x)=8x^2-x^3,(2,24) ?
The tangent of f(x)=8x^2-x^3,(2,24) is y=20x-16