What is the tangent of y=x-x^2,(-1,-2) ?

The tangent of y=x-x^2,(-1,-2) is y=3x+1

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tangent of y=x-x^2,(-1,-2)

tangent of

y = x - x^{2}, (- 1, - 2)

y = 3 x + 1

Solution steps

Find the slope of

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

Find the line with slope m=

3

and passing through

(- 1, - 2) : y = 3 x + 1

y = 3 x + 1

t^{3} sin (3 t)

f^{^{'}} (x) = 5 x + 1

(tan (y))^{^{'}}

x^{^{'}} = - 3 \cdot \frac{x}{300}

\int_{- 2}^{8} 6 x^{2} + 9 d x

What is the tangent of y=x-x^2,(-1,-2) ?
The tangent of y=x-x^2,(-1,-2) is y=3x+1