What is the tangent of f(x)=3x^2+12x-36 ?

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

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tangent of f(x)=3x^2+12x-36

tangent of

f (x) = 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

f (x) = 3 x^{2} + 12 x - 36 : \frac{df ( x )}{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

\int_{- 3}^{1} x^{2} + 2 x - 3 d x

\frac{d y}{d x} + y^{7} x + 2 y = 0

\int - \frac{10}{x ^{2} + 16} d x

\int f (u) d u

f (x) = \frac{5 x}{x - 2}, (3, 15)

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