What is the tangent of 3x^2-4x+1 ?

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

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tangent of 3x^2-4x+1

tangent of

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

y = 3 x^{2} - 4 x + 1 : \frac{d y}{d x} = 6 x - 4

Compute the slope of

y = 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

\frac{d}{d x} ((8 (3 x)^{4} + 5)^{3})

\int \frac{20 x ^{2} + x + 45}{x ( 4 x ^{2} + 9 )} d x

y = \frac{4}{( 2 x ) ^{3}}

\int (- 7 - 7 tan^{2} (θ)) d θ

\frac{d P}{d x} = 0.05 P (1 - \frac{P}{2000})

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