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

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

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

tangent of

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

y = - 4 x - 7

Solution steps

Find the tangent point:

(- 2, 1)

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

x = - 2 : 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}{2 x ^{2} + 5} d x

\int \frac{3 x + 1}{x ^{2} + x - 6} d x

x = 7 y^{2}, x = 3 + 4 y^{2}

\int sin (mπ x) d x

f (x) = 2 x - \frac{e ^{x}}{2}

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