What is the tangent of 6x^2+4x-3 ?

The tangent of 6x^2+4x-3 is y=(12a_{0}+4)x-6a_{0}^2-3

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tangent of 6x^2+4x-3

tangent of

6 x^{2} + 4 x - 3

y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 3

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} - 3)

Find the slope of

y = 6 x^{2} + 4 x - 3 : \frac{d y}{d x} = 12 x + 4

Compute the slope of

y = 6 x^{2} + 4 x - 3

x = a_{0} : 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} - 3) : y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 3

y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 3

\int \frac{x ^{n + 1}}{n + 1} d x

\frac{d}{d x} (ln^{x} (x))

\frac{d}{d t} (\frac{1}{e ^{t}})

f (x) = \frac{1 + 3 x}{1 - 3 x}

\int_{0}^{1} [x - x^{k}] d x

What is the tangent of 6x^2+4x-3 ?
The tangent of 6x^2+4x-3 is y=(12a_{0}+4)x-6a_{0}^2-3