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

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

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

tangent of

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

y = 2 x - 4

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, - 2) : y = 2 x - 4

y = 2 x - 4

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

n = 0 \sum \infty n + 1 (\frac{1}{9})^{n}

\frac{\partial}{\partial x} (3 x^{2} \cdot cos^{2} (x - y))

\int x (2 x^{2} - 4)^{3} d x

x \to 8 + lim (\frac{6 x}{x - 8})

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