What is the tangent of f(x)=9-x^2,\at x=-1 ?

The tangent of f(x)=9-x^2,\at x=-1 is y=2x+10

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tangent of f(x)=9-x^2,\at x=-1

tangent of

f (x) = 9 - x^{2}, at x = - 1

y = 2 x + 10

Solution steps

Find the tangent point:

(- 1, 8)

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}

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

(- 1, 8) : y = 2 x + 10

y = 2 x + 10

x^{- \frac{2}{3}}

\int \frac{7 x + 6}{x ^{2} ( x + 2 )} d x

y = 8 x^{2} + 8, (2, 40)

\frac{\partial}{\partial z} (x e^{y - z^{2}})

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

What is the tangent of f(x)=9-x^2,\at x=-1 ?
The tangent of f(x)=9-x^2,\at x=-1 is y=2x+10