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

The tangent of y=x^2-3,(-2,1) is y=-4x-7

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

tangent of

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

y = - 4 x - 7

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 3

(- 2, 1) : m = - 4

Find the line with slope m=

- 4

and passing through

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

y = - 4 x - 7

\int \frac{x ^{5} + 1}{x ^{3} - 3 x ^{2} - 10 x} d x

\frac{d}{d x} (e^{x} x + 7 e^{x})

\int (e^{x} - 3)^{2} d x

10 x^{4} - 10 x^{2}, 9 x^{2}

\frac{\partial}{\partial y} ((2 - x) (y - 2) (3 - x - y))

What is the tangent of y=x^2-3,(-2,1) ?
The tangent of y=x^2-3,(-2,1) is y=-4x-7