What is the tangent of f(x)=2x^3-x^2+9,(2,21) ?

The tangent of f(x)=2x^3-x^2+9,(2,21) is y=20x-19

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tangent of f(x)=2x^3-x^2+9,(2,21)

tangent of

f (x) = 2 x^{3} - x^{2} + 9, (2, 21)

y = 20 x - 19

Solution steps

Find the slope of

f (x) = 2 x^{3} - x^{2} + 9 : \frac{df ( x )}{d x} = 6 x^{2} - 2 x

Compute the slope of

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

(2, 21) : m = 20

Find the line with slope m=

20

and passing through

(2, 21) : y = 20 x - 19

y = 20 x - 19

\int \frac{4}{35 sin ( x ) - 12 cos ( x )} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e^{- t}

\int f x d x

\int (\frac{1}{5} - \frac{2}{x}) d x

x \to 8 lim (x^{\frac{7}{3}})

What is the tangent of f(x)=2x^3-x^2+9,(2,21) ?
The tangent of f(x)=2x^3-x^2+9,(2,21) is y=20x-19