What is the tangent of y=9+4x^2-2x^3,(1,11) ?

The tangent of y=9+4x^2-2x^3,(1,11) is y=2x+9

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tangent of y=9+4x^2-2x^3,(1,11)

tangent of

y = 9 + 4 x^{2} - 2 x^{3}, (1, 11)

y = 2 x + 9

Solution steps

Find the slope of

y = 9 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

Compute the slope of

y = 9 + 4 x^{2} - 2 x^{3}

(1, 11) : m = 2

Find the line with slope m=

2

and passing through

(1, 11) : y = 2 x + 9

y = 2 x + 9

\frac{d}{d x} (\frac{9}{ln ( x )})

\int - 4 t d t

f (x) = - 1 - x x^{2}

\int cos (22 x) cos (17 x) d x

n = 3 \sum \infty 5^{n}

What is the tangent of y=9+4x^2-2x^3,(1,11) ?
The tangent of y=9+4x^2-2x^3,(1,11) is y=2x+9