What is the tangent of y=x^3-16x^2+45x+6 ?

The tangent of y=x^3-16x^2+45x+6 is y=(3a_{0}^2-32a_{0}+45)x-2a_{0}^3+16a_{0}^2+6

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tangent of y=x^3-16x^2+45x+6

tangent of

y = x^{3} - 16 x^{2} + 45 x + 6

y = (3 a_{0}^{2} - 32 a_{0} + 45) x - 2 a_{0}^{3} + 16 a_{0}^{2} + 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 16 a_{0}^{2} + 45 a_{0} + 6)

Find the slope of

y = x^{3} - 16 x^{2} + 45 x + 6 : \frac{d y}{d x} = 3 x^{2} - 32 x + 45

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 = 3 a_{0}^{2} - 32 a_{0} + 45

Find the line with slope m=

3 a_{0}^{2} - 32 a_{0} + 45

and passing through

(a_{0}, a_{0}^{3} - 16 a_{0}^{2} + 45 a_{0} + 6) : y = (3 a_{0}^{2} - 32 a_{0} + 45) x - 2 a_{0}^{3} + 16 a_{0}^{2} + 6

y = (3 a_{0}^{2} - 32 a_{0} + 45) x - 2 a_{0}^{3} + 16 a_{0}^{2} + 6

\int 4 sin (x) + 4 d x

\frac{d y}{d x} = \frac{1}{4} y

x \to - 5 - lim (\frac{6}{x ^{2} - 25})

f (x) = sin (x^{2} + 2)

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

What is the tangent of y=x^3-16x^2+45x+6 ?
The tangent of y=x^3-16x^2+45x+6 is y=(3a_{0}^2-32a_{0}+45)x-2a_{0}^3+16a_{0}^2+6