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

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

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

tangent of

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

y = - 2 x - 1

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x - 1

\frac{d}{d x} (- 5 sin (x))

\frac{\partial}{\partial x} (2 - x^{2} - y^{2})

\frac{d}{d x} (- e^{- 4 x})

\int_{3}^{\infty} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

(cos (x))^{^{'}}

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