What is the tangent of y=2x^4-16x^2+9,\at ?

The tangent of y=2x^4-16x^2+9,\at is y=(8a_{0}^3-32a_{0})x-6a_{0}^4+16a_{0}^2+9

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tangent of y=2x^4-16x^2+9,\at

tangent of

y = 2 x^{4} - 16 x^{2} + 9, at

y = (8 a_{0}^{3} - 32 a_{0}) x - 6 a_{0}^{4} + 16 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{4} - 16 a_{0}^{2} + 9)

Find the slope of

y = 2 x^{4} - 16 x^{2} + 9 : \frac{d y}{d x} = 8 x^{3} - 32 x

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 = 8 a_{0}^{3} - 32 a_{0}

Find the line with slope m=

8 a_{0}^{3} - 32 a_{0}

and passing through

(a_{0}, 2 a_{0}^{4} - 16 a_{0}^{2} + 9) : y = (8 a_{0}^{3} - 32 a_{0}) x - 6 a_{0}^{4} + 16 a_{0}^{2} + 9

y = (8 a_{0}^{3} - 32 a_{0}) x - 6 a_{0}^{4} + 16 a_{0}^{2} + 9

\int \frac{x}{3 e ^{2 x}} d x

x \cdot \frac{d y}{d x} - y = 5 x^{2}

h \to 0 lim (2 h)

\int 4 x - 8 y^{3}

\frac{d y}{d x} = 2 x + y - 2

What is the tangent of y=2x^4-16x^2+9,\at ?
The tangent of y=2x^4-16x^2+9,\at is y=(8a_{0}^3-32a_{0})x-6a_{0}^4+16a_{0}^2+9