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

The tangent of f(x)=6x^2-4x+4,(1,6) is y=8x-2

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tangent of f(x)=6x^2-4x+4,(1,6)

tangent of

f (x) = 6 x^{2} - 4 x + 4, (1, 6)

y = 8 x - 2

Solution steps

Find the slope of

f (x) = 6 x^{2} - 4 x + 4 : \frac{df ( x )}{d x} = 12 x - 4

Compute the slope of

f (x) = 6 x^{2} - 4 x + 4

(1, 6) : m = 8

Find the line with slope m=

8

and passing through

(1, 6) : y = 8 x - 2

y = 8 x - 2

\frac{\partial}{\partial y} (4 - x^{2} - 8 y^{2})

y = \frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (3 x, - 2 y, x - y)

\frac{d}{d x} ((e^{- x^{2}}) (- 2 x))

y^{^{''}} + y = - sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

What is the tangent of f(x)=6x^2-4x+4,(1,6) ?
The tangent of f(x)=6x^2-4x+4,(1,6) is y=8x-2