What is the tangent of f(x)=9x^2-x^3,(2,28) ?

The tangent of f(x)=9x^2-x^3,(2,28) is y=24x-20

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tangent of f(x)=9x^2-x^3,(2,28)

tangent of

f (x) = 9 x^{2} - x^{3}, (2, 28)

y = 24 x - 20

Solution steps

Find the slope of

f (x) = 9 x^{2} - x^{3} : \frac{df ( x )}{d x} = 18 x - 3 x^{2}

Compute the slope of

f (x) = 9 x^{2} - x^{3}

(2, 28) : m = 24

Find the line with slope m=

24

and passing through

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

y = 24 x - 20

\frac{d}{d x} (\frac{1}{3 x ^{5}})

f (x) = (\frac{sin ( x )}{x})^{2}

\int 4 x^{\frac{2}{3}} d x

\frac{\partial}{\partial y} (3 x + 2 x y^{2})

\int \frac{4}{( x ^{2} + 5 x - 14 )} d x

What is the tangent of f(x)=9x^2-x^3,(2,28) ?
The tangent of f(x)=9x^2-x^3,(2,28) is y=24x-20