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

Compute the slope of

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

x = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x + 1

\int 5^{2} - x^{2} d x

\frac{\partial}{\partial x} (x tan (x))

y = 3 x (a^{2} - x^{2})

4 (1 + e^{- 0.3 t})

\int \frac{( ln ( x ) ) ^{3}}{2 x} d x

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