What is the tangent of f(x)=3x^2-4x+1 ?

The tangent of f(x)=3x^2-4x+1 is y=(6a_{0}-4)x-3a_{0}^2+1

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tangent of f(x)=3x^2-4x+1

tangent of

f (x) = 3 x^{2} - 4 x + 1

y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 4 a_{0} + 1)

Find the slope of

f (x) = 3 x^{2} - 4 x + 1 : \frac{df ( x )}{d x} = 6 x - 4

Compute the slope of

f (x) = 3 x^{2} - 4 x + 1

x = a_{0} : m = 6 a_{0} - 4

Find the line with slope m=

6 a_{0} - 4

and passing through

(a_{0}, 3 a_{0}^{2} - 4 a_{0} + 1) : y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 1

y = (6 a_{0} - 4) x - 3 a_{0}^{2} + 1

y = 4, y = x^{2}

\int 2 e^{- \frac{x}{2}} d x

\frac{d}{d x} (\frac{9 - sec ( x )}{tan ( x )})

n = 1 \sum \infty \frac{2 n}{5 n - 8}

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

What is the tangent of f(x)=3x^2-4x+1 ?
The tangent of f(x)=3x^2-4x+1 is y=(6a_{0}-4)x-3a_{0}^2+1