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

The tangent of f(x)=4x^2-4x+1,\at x=1,y=0 is y=4x-3

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

tangent of

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

y = 4 x - 3

Solution steps

Find the tangent point:

(1, 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 = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 1) : y = 4 x - 3

y = 4 x - 3

\int cos (4 x) sin (x) d x

y = \frac{x ^{2} + 1}{x ^{3} - 1} - 5 x^{2}

\frac{e ^{x}}{1 + e ^{x}}

x^{3} + 4 y^{2} = 12, (2, - 1)

\frac{d y}{d x} = x - \frac{y}{x}

What is the tangent of f(x)=4x^2-4x+1,\at x=1,y=0 ?
The tangent of f(x)=4x^2-4x+1,\at x=1,y=0 is y=4x-3