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

The tangent of f(x)=4x^2-2x+1 is y=(8a_{0}-2)x-4a_{0}^2+1

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

tangent of

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

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 2 a_{0} + 1)

Find the slope of

f (x) = 4 x^{2} - 2 x + 1 : \frac{df ( x )}{d x} = 8 x - 2

Compute the slope of

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

x = a_{0} : m = 8 a_{0} - 2

Find the line with slope m=

8 a_{0} - 2

and passing through

(a_{0}, 4 a_{0}^{2} - 2 a_{0} + 1) : y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 1

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 1

\int \frac{5 x ^{4}}{1 + x ^{5}} d x

f (x) = 2 x + 1, at x = 3

g (t) = tan (cos (2 t))

(- 1, - 1), (1, 5)

\frac{d}{d x} ((3 x - 1)^{4} (2 x + 1)^{- 3})

What is the tangent of f(x)=4x^2-2x+1 ?
The tangent of f(x)=4x^2-2x+1 is y=(8a_{0}-2)x-4a_{0}^2+1