What is the tangent of 6x^2-4x+5 ?

The tangent of 6x^2-4x+5 is y=(12a_{0}-4)x-6a_{0}^2+5

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tangent of 6x^2-4x+5

tangent of

6 x^{2} - 4 x + 5

y = (12 a_{0} - 4) x - 6 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} - 4 a_{0} + 5)

Find the slope of

y = 6 x^{2} - 4 x + 5 : \frac{d y}{d x} = 12 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 = 12 a_{0} - 4

Find the line with slope m=

12 a_{0} - 4

and passing through

(a_{0}, 6 a_{0}^{2} - 4 a_{0} + 5) : y = (12 a_{0} - 4) x - 6 a_{0}^{2} + 5

y = (12 a_{0} - 4) x - 6 a_{0}^{2} + 5

\int \frac{( x ^{4} - 5 x ^{2} + 3 x - 4 )}{x} d x

\int_{1}^{e} (\frac{x ^{2} - 1}{x}) d x

\frac{d}{d x} (ln (\frac{x + y}{x - y}))

\frac{\partial}{\partial s} (sin (s^{2} - t^{2}))

\frac{d}{d x} (ln (2 x^{4} e^{3 x} (x^{2} + x + 1)^{5}))

What is the tangent of 6x^2-4x+5 ?
The tangent of 6x^2-4x+5 is y=(12a_{0}-4)x-6a_{0}^2+5