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

\frac{d x}{d t} = - 40.8 x + 0.6

x \to - \infty lim ((2)^{x})

x \to \infty lim (1, x)

\int_{0}^{4} x 21 - x^{2} d x

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 - e^{- 2 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