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

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

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tangent of y=-3x^2+2

tangent of

y = - 3 x^{2} + 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

- 6 a_{0}

and passing through

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

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

x \to 0 lim (\frac{5 x}{tan ( x )})

\int 2 + tan^{2} (θ) d θ

\frac{d}{d x} (\frac{x ^{3}}{7})

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

arcsin (x)

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