What is the tangent of y=x^3-9x,(3,0) ?

The tangent of y=x^3-9x,(3,0) is y=18x-54

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tangent of y=x^3-9x,(3,0)

tangent of

y = x^{3} - 9 x, (3, 0)

y = 18 x - 54

Solution steps

Find the slope of

y = x^{3} - 9 x : \frac{d y}{d x} = 3 x^{2} - 9

Compute the slope of

y = x^{3} - 9 x

(3, 0) : m = 18

Find the line with slope m=

18

and passing through

(3, 0) : y = 18 x - 54

y = 18 x - 54

f (x) = x - 5

x \to \infty lim (- \frac{3}{x})

y^{^{'}} + 3 (tan (3 x)) y = - 2 cos (3 x)

\int_{0}^{7} \frac{5 t}{( t - 8 ) ^{2}} d t

\frac{\partial}{\partial x} (e^{(- 0.1 x y)})

What is the tangent of y=x^3-9x,(3,0) ?
The tangent of y=x^3-9x,(3,0) is y=18x-54