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

The tangent of f(x)=x^2-9x,\at x=3 is y=-3x-9

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

tangent of

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

y = - 3 x - 9

Solution steps

Find the tangent point:

(3, - 18)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 3

Find the line with slope m=

- 3

and passing through

(3, - 18) : y = - 3 x - 9

y = - 3 x - 9

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

\frac{1}{( z + 1 ) ^{3}}

\frac{d}{d x} (x^{2} + x + 3)

(x^{4} + y^{2}) d x - x y d y = 0

\int (tan (θ)) d θ

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