What is the tangent of f(x)=4x^2-4x+1,\at x=0 ?

The tangent of f(x)=4x^2-4x+1,\at x=0 is y=-4x+1

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tangent of f(x)=4x^2-4x+1,\at x=0

tangent of

f (x) = 4 x^{2} - 4 x + 1, at x = 0

y = - 4 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

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 = - 4

Find the line with slope m=

- 4

and passing through

(0, 1) : y = - 4 x + 1

y = - 4 x + 1

What is the tangent of f(x)=4x^2-4x+1,\at x=0 ?
The tangent of f(x)=4x^2-4x+1,\at x=0 is y=-4x+1