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

The tangent of f(x)=1-x^2,\at x=2 is y=-4x+5

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

tangent of

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

y = - 4 x + 5

Solution steps

Find the tangent point:

(2, - 3)

Find the slope of

f (x) = 1 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 1 - x^{2}

x = 2 : m = - 4

Find the line with slope m=

- 4

and passing through

(2, - 3) : y = - 4 x + 5

y = - 4 x + 5

\int \frac{21}{25 + 4 r ^{2}} d r

\frac{1}{x} - 1

\frac{\partial}{\partial x} (2 y - 3 x + 5)

\int (e^{6 x} + e^{- 9 x}) d x

\int x^{2} \cdot cos (n x) d x

What is the tangent of f(x)=1-x^2,\at x=2 ?
The tangent of f(x)=1-x^2,\at x=2 is y=-4x+5