What is the tangent of f(x)=12-x^2,(3,3) ?

The tangent of f(x)=12-x^2,(3,3) is y=-6x+21

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tangent of f(x)=12-x^2,(3,3)

tangent of

f (x) = 12 - x^{2}, (3, 3)

y = - 6 x + 21

Solution steps

Find the slope of

f (x) = 12 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 12 - x^{2}

(3, 3) : m = - 6

Find the line with slope m=

- 6

and passing through

(3, 3) : y = - 6 x + 21

y = - 6 x + 21

f (x) = \frac{18}{x ^{2} + 2}, (1, 6)

x \to 0 + lim (x^{11 x})

\int e^{- 6 x} 9 x d x

\frac{d}{d x} (5 x^{5} sin (2 x))

\frac{d}{d x} (e^{(0.5 x)})

What is the tangent of f(x)=12-x^2,(3,3) ?
The tangent of f(x)=12-x^2,(3,3) is y=-6x+21