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

The tangent of f(x)=5x^2-4x-2,\at x=1 is y=6x-7

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

tangent of

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

y = 6 x - 7

Solution steps

Find the tangent point:

(1, - 1)

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 : m = 6

Find the line with slope m=

6

and passing through

(1, - 1) : y = 6 x - 7

y = 6 x - 7

t y^{^{'}} + 10 y = \frac{1}{t}

\int 7 \frac{cos ( x ) + sin ( 2 x )}{sin ( x )} d x

1 - e^{- 2 x}

\int \frac{1}{10 x - 7} d x

x \to 8 lim (e (\frac{3}{( x - 8 )}))

What is the tangent of f(x)=5x^2-4x-2,\at x=1 ?
The tangent of f(x)=5x^2-4x-2,\at x=1 is y=6x-7