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

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

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tangent of f(x)=-x^2

tangent of

f (x) = - x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = - x^{2}

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

Find the line with slope m=

- 2 a_{0}

and passing through

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

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

y^{^{'}} - 6 y = 5

\int \frac{5 x - 7}{( x ^{2} - x - 2 ) ( x - 3 )} d x

\frac{d}{d x} (6 ln (x) + 11)

\int_{- 0.5}^{1} cos (x^{2}) d x

\frac{d y}{d x} = \frac{y}{x} + 6 x + 7

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