What is the tangent of y=-x^3+2x^2+1,\at x=-1 ?

The tangent of y=-x^3+2x^2+1,\at x=-1 is y=-7x-3

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tangent of y=-x^3+2x^2+1,\at x=-1

tangent of

y = - x^{3} + 2 x^{2} + 1, at x = - 1

y = - 7 x - 3

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

y = - x^{3} + 2 x^{2} + 1 : \frac{d y}{d x} = - 3 x^{2} + 4 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 = - 7

Find the line with slope m=

- 7

and passing through

(- 1, 4) : y = - 7 x - 3

y = - 7 x - 3

y^{^{''}} - 2 a y^{^{'}} + a^{2} y = 0

\int - sin (θ) d θ

\int 3 t^{2} - 2 t d t

\int_{- 1}^{2} 4 x (1 - x^{2}) d x

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

What is the tangent of y=-x^3+2x^2+1,\at x=-1 ?
The tangent of y=-x^3+2x^2+1,\at x=-1 is y=-7x-3