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

The tangent of y=x^2,(-1,-3) is y=2x-1 \lor y=-6x-9

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

tangent of

y = x^{2}, (- 1, - 3)

y = 2 x - 1 or y = - 6 x - 9

Solution steps

Slope of the tangent line at

(- 1, - 3) : m = \frac{x ^{2} + 3}{x + 1}

Find the slope of

y = x^{2} : m = 2 x

Solve

2 x = \frac{x ^{2} + 3}{x + 1} : x = 1, x = - 3

Find the tangent line at

x = 1 : y = 2 x - 1

Find the tangent line at

x = - 3 : y = - 6 x - 9

y = 2 x - 1 or y = - 6 x - 9

\int 5 e^{- 2 x} d x

\int_{v^{1}}^{v^{2}} \frac{1}{v} d v

x \to - 1 lim (\frac{1}{x ^{2} - 1})

\frac{d ^{2}}{d x ^{2}} (9 x sin (x^{2}))

\frac{\partial}{\partial x} (2 x + 1)

What is the tangent of y=x^2,(-1,-3) ?
The tangent of y=x^2,(-1,-3) is y=2x-1 \lor y=-6x-9