What is the tangent of f(x)=x^2-2x,\at x=-3 ?

The tangent of f(x)=x^2-2x,\at x=-3 is y=-8x-9

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tangent of f(x)=x^2-2x,\at x=-3

tangent of

f (x) = x^{2} - 2 x, at x = - 3

y = - 8 x - 9

Solution steps

Find the tangent point:

(- 3, 15)

Find the slope of

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

Compute the slope of

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

x = - 3 : m = - 8

Find the line with slope m=

- 8

and passing through

(- 3, 15) : y = - 8 x - 9

y = - 8 x - 9

\frac{d}{d x} (\frac{- x ^{2} - 2 x + 1}{( x ^{2} + 1 ) ^{2}})

x \to 2 - lim (x^{2} - x + 2)

n = 1 \sum \infty \frac{( 2 ^{n} )}{n !}

\frac{\partial}{\partial x} (2 x ln (x y))

\int_{0}^{50} 0.624 x d x

What is the tangent of f(x)=x^2-2x,\at x=-3 ?
The tangent of f(x)=x^2-2x,\at x=-3 is y=-8x-9