What is the tangent of f(x)=x^2-2,(-4,14) ?

The tangent of f(x)=x^2-2,(-4,14) is y=-8x-18

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tangent of f(x)=x^2-2,(-4,14)

tangent of

f (x) = x^{2} - 2, (- 4, 14)

y = - 8 x - 18

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 2

(- 4, 14) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 14) : y = - 8 x - 18

y = - 8 x - 18

\frac{d}{d x} (\frac{2 x - 8}{4 x - 3})

\frac{d}{d x} (\frac{cos ( 4 x )}{x})

\frac{\partial}{\partial x} (sin (y e^{x y}))

\int e^{y x} d x

y^{^{''}} + 9 y^{^{'}} = 0, y (0) = 1

What is the tangent of f(x)=x^2-2,(-4,14) ?
The tangent of f(x)=x^2-2,(-4,14) is y=-8x-18