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

The tangent of f(x)=5-x^2,\at x=-1 is y=2x+6

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tangent of f(x)=5-x^2,\at x=-1

tangent of

f (x) = 5 - x^{2}, at x = - 1

y = 2 x + 6

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

f (x) = 5 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 5 - x^{2}

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

(- 1, 4) : y = 2 x + 6

y = 2 x + 6

\int \frac{1}{x ^{6} - x ^{3}} d x

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

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}

\frac{2}{u}

\int (x^{2} - 9 x + 3) d x

What is the tangent of f(x)=5-x^2,\at x=-1 ?
The tangent of f(x)=5-x^2,\at x=-1 is y=2x+6