What is the tangent of 3(x)^2+8(x)+12 ?

The tangent of 3(x)^2+8(x)+12 is y=(6a_{0}+8)x-3a_{0}^2+12

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tangent of 3(x)^2+8(x)+12

tangent of

3 (x)^{2} + 8 (x) + 12

y = (6 a_{0} + 8) x - 3 a_{0}^{2} + 12

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 8 a_{0} + 12)

Find the slope of

y = 3 x^{2} + 8 x + 12 : \frac{d y}{d x} = 6 x + 8

Compute the slope of

y = 3 x^{2} + 8 x + 12

x = a_{0} : m = 6 a_{0} + 8

Find the line with slope m=

6 a_{0} + 8

and passing through

(a_{0}, 3 a_{0}^{2} + 8 a_{0} + 12) : y = (6 a_{0} + 8) x - 3 a_{0}^{2} + 12

y = (6 a_{0} + 8) x - 3 a_{0}^{2} + 12

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

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

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

x \to 2 lim (3 x + 2)

y = 6 x - x^{2}

What is the tangent of 3(x)^2+8(x)+12 ?
The tangent of 3(x)^2+8(x)+12 is y=(6a_{0}+8)x-3a_{0}^2+12