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

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

t an g e n t f (x) = 2 x + 1, at x = 3

s l o p e in t erce pt (- 1, - 1), (1, 5)

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

x y^{^{'}} - y = x^{2} + 5 x - 1

t an g e n t f (x) = 3 x^{5} - 2 x^{4}, at x = 4

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