What is the tangent of-3x^2-8x ?

The tangent of-3x^2-8x is y=(-6a_{0}-8)x+3a_{0}^2

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tangent of-3x^2-8x

tangent of

- 3 x^{2} - 8 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = - 3 x^{2} - 8 x : \frac{d y}{d x} = - 6 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 = - 6 a_{0} - 8

Find the line with slope m=

- 6 a_{0} - 8

and passing through

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

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

y^{^{''}} + 8 y^{^{'}} + 16 y = 0

\frac{d y}{d x} = x^{5} y

\frac{d}{d t} (\frac{t ^{2} + 2}{t ^{4} - 3 t ^{2} + 1})

\frac{d}{d x} (4 sin^{2} (2 x))

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 4)

What is the tangent of-3x^2-8x ?
The tangent of-3x^2-8x is y=(-6a_{0}-8)x+3a_{0}^2