What is the tangent of f(x)=x^2,(-0.8,0.64) ?

The tangent of f(x)=x^2,(-0.8,0.64) is y=-1.6x-0.64

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tangent of f(x)=x^2,(-0.8,0.64)

tangent of

f (x) = x^{2}, (- 0.8, 0.64)

y = - 1.6 x - 0.64

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2}

(- 0.8, 0.64) : m = - 1.6

Find the line with slope m=

- 1.6

and passing through

(- 0.8, 0.64) : y = - 1.6 x - 0.64

y = - 1.6 x - 0.64

\int \frac{5 x}{( x ^{2} + 6 ) ^{2}} d x

\int 0 e^{x} d x

y^{^{''}} sin (x) cos (x) - 2 y^{^{'}} = 0

\int 6 e^{- 5 x} d x

x \to 0 lim (\frac{2}{sin ( x )})

What is the tangent of f(x)=x^2,(-0.8,0.64) ?
The tangent of f(x)=x^2,(-0.8,0.64) is y=-1.6x-0.64