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

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

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

tangent of

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

y = - 4 x - 7

Solution steps

Find the slope of

f (x) = x^{2} - 3 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = 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{1}{x ^{\frac{1}{2}} - 4} d x

\int x^{2} 5 - 2 x^{3} d x

\int (x^{1.8} + 11 x^{4.5}) d x

\frac{\partial}{\partial y} (7 e^{x} ln (y))

x \to 6 lim (x (x - 3))

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