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

The tangent of f(x)=-x^2,(-3,-9) is y=6x+9

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

tangent of

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

y = 6 x + 9

Solution steps

Find the slope of

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

Compute the slope of

f (x) = - x^{2}

(- 3, - 9) : m = 6

Find the line with slope m=

6

and passing through

(- 3, - 9) : y = 6 x + 9

y = 6 x + 9

\int \frac{2}{( x - 2 ) ( x ^{4} + 4 )} d x

x \to - \infty lim (- 3 - x)

\int_{0}^{3} x^{2} + 16 d x

\frac{d y}{d x} + \frac{y}{x} = \frac{- y ^{2}}{x}

\int \frac{4 x}{( 2 x ^{2} + 3 ) ^{3}} d x

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