What is the tangent of f(x)=-x^3+9x^2-6x+9,\at x=7 ?

The tangent of f(x)=-x^3+9x^2-6x+9,\at x=7 is y=-27x+254

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tangent of f(x)=-x^3+9x^2-6x+9,\at x=7

tangent of

f (x) = - x^{3} + 9 x^{2} - 6 x + 9, at x = 7

y = - 27 x + 254

Solution steps

Find the tangent point:

(7, 65)

Find the slope of

f (x) = - x^{3} + 9 x^{2} - 6 x + 9 : \frac{df ( x )}{d x} = - 3 x^{2} + 18 x - 6

Compute the slope of

f (x) = - x^{3} + 9 x^{2} - 6 x + 9

x = 7 : m = - 27

Find the line with slope m=

- 27

and passing through

(7, 65) : y = - 27 x + 254

y = - 27 x + 254

x \to - \infty lim (\frac{x}{c})

\int 2 (2 x + 8)^{5} d x

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

\int_{0}^{6} \frac{1}{36 - x ^{2}} d x

\int (y + 2) d y

What is the tangent of f(x)=-x^3+9x^2-6x+9,\at x=7 ?
The tangent of f(x)=-x^3+9x^2-6x+9,\at x=7 is y=-27x+254