What is the tangent of f(x)=x^2-6x+1,\at x=1 ?

The tangent of f(x)=x^2-6x+1,\at x=1 is y=-4x

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tangent of f(x)=x^2-6x+1,\at x=1

tangent of

f (x) = x^{2} - 6 x + 1, at x = 1

y = - 4 x

Solution steps

Find the tangent point:

(1, - 4)

Find the slope of

f (x) = x^{2} - 6 x + 1 : \frac{df ( x )}{d x} = 2 x - 6

Compute the slope of

f (x) = x^{2} - 6 x + 1

x = 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(1, - 4) : y = - 4 x

y = - 4 x

f (x) = \frac{1}{ln ( x )}

x \to 5 lim (\frac{\frac{5}{x} - 1}{x - 5})

\frac{\partial}{\partial y} (6 x sin (y - z))

x^{3} cos (2 x)

What is the tangent of f(x)=x^2-6x+1,\at x=1 ?
The tangent of f(x)=x^2-6x+1,\at x=1 is y=-4x