What is the tangent of 2x^2+8x+9 ?

The tangent of 2x^2+8x+9 is y=(4a_{0}+8)x-2a_{0}^2+9

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tangent of 2x^2+8x+9

tangent of

2 x^{2} + 8 x + 9

y = (4 a_{0} + 8) x - 2 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 8 a_{0} + 9)

Find the slope of

y = 2 x^{2} + 8 x + 9 : \frac{d y}{d x} = 4 x + 8

Compute the slope of

y = 2 x^{2} + 8 x + 9

x = a_{0} : m = 4 a_{0} + 8

Find the line with slope m=

4 a_{0} + 8

and passing through

(a_{0}, 2 a_{0}^{2} + 8 a_{0} + 9) : y = (4 a_{0} + 8) x - 2 a_{0}^{2} + 9

y = (4 a_{0} + 8) x - 2 a_{0}^{2} + 9

x \to 2 + lim (\frac{x - 1}{x - 2})

\int_{1}^{t} 15 x^{- 3} d x

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

\int sec (π x) tan (π x) d x

\int 8 e^{8 x - 3} d x

What is the tangent of 2x^2+8x+9 ?
The tangent of 2x^2+8x+9 is y=(4a_{0}+8)x-2a_{0}^2+9