What is the tangent of f(x)=2x^3+42x^2+270x+4 ?

The tangent of f(x)=2x^3+42x^2+270x+4 is y=(6a_{0}^2+84a_{0}+270)x-4a_{0}^3-42a_{0}^2+4

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tangent of f(x)=2x^3+42x^2+270x+4

tangent of

f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4

y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 270 a_{0} + 4)

Find the slope of

f (x) = 2 x^{3} + 42 x^{2} + 270 x + 4 : \frac{df ( x )}{d x} = 6 x^{2} + 84 x + 270

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}^{2} + 84 a_{0} + 270

Find the line with slope m=

6 a_{0}^{2} + 84 a_{0} + 270

and passing through

(a_{0}, 2 a_{0}^{3} + 42 a_{0}^{2} + 270 a_{0} + 4) : y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

y = (6 a_{0}^{2} + 84 a_{0} + 270) x - 4 a_{0}^{3} - 42 a_{0}^{2} + 4

\int sin (x) sec^{8} (x) d x

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

\frac{d}{d x} (9 tan (x) sec (x))

\int \frac{1}{1 + 3 x} d x

\frac{5}{4 + x ^{2}}, at x = 2

What is the tangent of f(x)=2x^3+42x^2+270x+4 ?
The tangent of f(x)=2x^3+42x^2+270x+4 is y=(6a_{0}^2+84a_{0}+270)x-4a_{0}^3-42a_{0}^2+4