What is the tangent of y=x^3-7x^2+8x+1 ?

The tangent of y=x^3-7x^2+8x+1 is y=(3a_{0}^2-14a_{0}+8)x-2a_{0}^3+7a_{0}^2+1

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tangent of y=x^3-7x^2+8x+1

tangent of

y = x^{3} - 7 x^{2} + 8 x + 1

y = (3 a_{0}^{2} - 14 a_{0} + 8) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 7 a_{0}^{2} + 8 a_{0} + 1)

Find the slope of

y = x^{3} - 7 x^{2} + 8 x + 1 : \frac{d y}{d x} = 3 x^{2} - 14 x + 8

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} - 14 a_{0} + 8

Find the line with slope m=

3 a_{0}^{2} - 14 a_{0} + 8

and passing through

(a_{0}, a_{0}^{3} - 7 a_{0}^{2} + 8 a_{0} + 1) : y = (3 a_{0}^{2} - 14 a_{0} + 8) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 1

y = (3 a_{0}^{2} - 14 a_{0} + 8) x - 2 a_{0}^{3} + 7 a_{0}^{2} + 1

\int \frac{64}{x ^{3} - 8 x ^{2}} d x

\frac{d}{d x} (\frac{3 x ^{2} - x - 1}{3 x})

2 y^{^{''}} + y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{\partial}{\partial y} (x \cdot e^{x \cdot y})

\frac{\partial}{\partial x} (- \frac{x}{( x ^{2} + y ^{2} )})

What is the tangent of y=x^3-7x^2+8x+1 ?
The tangent of y=x^3-7x^2+8x+1 is y=(3a_{0}^2-14a_{0}+8)x-2a_{0}^3+7a_{0}^2+1