What is the tangent of 3x^2+12x-36 ?

The tangent of 3x^2+12x-36 is y=(6a_{0}+12)x-3a_{0}^2-36

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tangent of 3x^2+12x-36

tangent of

3 x^{2} + 12 x - 36

y = (6 a_{0} + 12) x - 3 a_{0}^{2} - 36

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 12 a_{0} - 36)

Find the slope of

y = 3 x^{2} + 12 x - 36 : \frac{d y}{d x} = 6 x + 12

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 = 6 a_{0} + 12

Find the line with slope m=

6 a_{0} + 12

and passing through

(a_{0}, 3 a_{0}^{2} + 12 a_{0} - 36) : y = (6 a_{0} + 12) x - 3 a_{0}^{2} - 36

y = (6 a_{0} + 12) x - 3 a_{0}^{2} - 36

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

\frac{d}{d x} (5 x^{2})

n = 0 \sum \infty tan (\frac{1}{n})

\int \frac{1}{( e ^{x} + e ^{- x} )} d x

x \to 10 lim (3)

What is the tangent of 3x^2+12x-36 ?
The tangent of 3x^2+12x-36 is y=(6a_{0}+12)x-3a_{0}^2-36