What is the tangent of f(x)=-4-x^2 ?

The tangent of f(x)=-4-x^2 is y=-2a_{0}x-4+a_{0}^2

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tangent of f(x)=-4-x^2

tangent of

f (x) = - 4 - x^{2}

y = - 2 a_{0} x - 4 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 4 - a_{0}^{2})

Find the slope of

f (x) = - 4 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = - 4 - x^{2}

x = a_{0} : m = - 2 a_{0}

Find the line with slope m=

- 2 a_{0}

and passing through

(a_{0}, - 4 - a_{0}^{2}) : y = - 2 a_{0} x - 4 + a_{0}^{2}

y = - 2 a_{0} x - 4 + a_{0}^{2}

f (x) = ln (3 + x)

y^{^{'}} = y + 1

\int_{3}^{3} \frac{1}{9 + x ^{2}} d x

\int_{- \infty}^{0} \frac{1}{3 - 9 x} d x

\frac{d}{d x} ((f (x))^{2})

What is the tangent of f(x)=-4-x^2 ?
The tangent of f(x)=-4-x^2 is y=-2a_{0}x-4+a_{0}^2