What is the tangent of f(x)=5x^2-4x-2,\at x=(1-1) ?

The tangent of f(x)=5x^2-4x-2,\at x=(1-1) is y=-4x-2

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tangent of f(x)=5x^2-4x-2,\at x=(1-1)

tangent of

f (x) = 5 x^{2} - 4 x - 2, at x = (1 - 1)

y = - 4 x - 2

Solution steps

Find the tangent point:

((1 - 1), - 2)

Find the slope of

f (x) = 5 x^{2} - 4 x - 2 : \frac{df ( x )}{d x} = 10 x - 4

Compute the slope of

f (x) = 5 x^{2} - 4 x - 2

x = (1 - 1) : m = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x - 2

\int_{- 3}^{0} \frac{x}{9 - x ^{2}} d x

y = x, y = \frac{1}{3} x, x = 25

\int \frac{x ^{2} + 16}{2 x} d x

\int_{0}^{8} \frac{1}{( 10 - 2 x ) ^{\frac{1}{5}}} d x

\int_{0}^{\infty} e^{- x} d x

What is the tangent of f(x)=5x^2-4x-2,\at x=(1-1) ?
The tangent of f(x)=5x^2-4x-2,\at x=(1-1) is y=-4x-2