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

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

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

tangent of

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

y = - 4 x + 1

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x + 1

f (x) = - 10 x + 5, (1, 2)

\int x^{2} 4 x^{2} + 1 d x

x \to 3 lim (\frac{x ^{2}}{x})

\frac{d}{d x} (\frac{sin ( 3 x + 1 )}{7 x ^{2} + 1})

\int \frac{12 x}{1 - x ^{4}} d x

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