What is the tangent of f(x)=9-x^2,(3,0) ?

The tangent of f(x)=9-x^2,(3,0) is y=-6x+18

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tangent of f(x)=9-x^2,(3,0)

tangent of

f (x) = 9 - x^{2}, (3, 0)

y = - 6 x + 18

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 0) : m = - 6

Find the line with slope m=

- 6

and passing through

(3, 0) : y = - 6 x + 18

y = - 6 x + 18

\frac{d}{d y} (y \cdot e^{x \cdot y})

y^{^{'}} = y + t

\int_{0}^{π} sin^{2} (x co) s^{2} x d x

y = x^{2} - x + 5, at x = 5

\int \frac{2 x + 1}{x ^{2} - 2 x + 5} d x

What is the tangent of f(x)=9-x^2,(3,0) ?
The tangent of f(x)=9-x^2,(3,0) is y=-6x+18