What is the tangent of 3x^2-26x+8 ?

The tangent of 3x^2-26x+8 is y=(6a_{0}-26)x-3a_{0}^2+8

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tangent of 3x^2-26x+8

tangent of

3 x^{2} - 26 x + 8

y = (6 a_{0} - 26) x - 3 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 26 a_{0} + 8)

Find the slope of

y = 3 x^{2} - 26 x + 8 : \frac{d y}{d x} = 6 x - 26

Compute the slope of

y = 3 x^{2} - 26 x + 8

x = a_{0} : m = 6 a_{0} - 26

Find the line with slope m=

6 a_{0} - 26

and passing through

(a_{0}, 3 a_{0}^{2} - 26 a_{0} + 8) : y = (6 a_{0} - 26) x - 3 a_{0}^{2} + 8

y = (6 a_{0} - 26) x - 3 a_{0}^{2} + 8

f (x) = 2 - 3 x

y^{^{'}} - 3 y = x y^{- 4}

\int tan^{4} (x) sec^{2} (x) d x

f (x) = x^{2} - 6 x, at x = - 2

What is the tangent of 3x^2-26x+8 ?
The tangent of 3x^2-26x+8 is y=(6a_{0}-26)x-3a_{0}^2+8