What is the tangent of f(x)=-1/4 x^2,(-2,-1) ?

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

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tangent of f(x)=-1/4 x^2,(-2,-1)

tangent of

f (x) = - \frac{1}{4} x^{2}, (- 2, - 1)

y = x + 1

Solution steps

Find the slope of

f (x) = - \frac{1}{4} x^{2} : \frac{df ( x )}{d x} = - \frac{x}{2}

Compute the slope of

f (x) = - \frac{1}{4} x^{2}

(- 2, - 1) : m = 1

Find the line with slope m=

1

and passing through

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

y = x + 1

\frac{1}{( s - 1 ) ^{3}}

\frac{d}{d x} ((8 x + e^{- x^{2}})^{3})

f (x) = \frac{6}{x ^{3}}

\frac{d y}{d x} = \frac{11 x}{y}

\frac{d}{d x} (arctan (- \frac{x}{2}))

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