What is the tangent of f(x)=x^2+2,(-2,6) ?

The tangent of f(x)=x^2+2,(-2,6) is y=-4x-2

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tangent of f(x)=x^2+2,(-2,6)

tangent of

f (x) = x^{2} + 2, (- 2, 6)

y = - 4 x - 2

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

(- 2, 6) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 6) : y = - 4 x - 2

y = - 4 x - 2

\int 2^{x - 1} d x

(2) (e^{3 x x})

\int (x + 1) e^{8 x^{2} + 16 x} d x

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

x \to 5 lim (\frac{x - 5}{x ^{2} - 5})

What is the tangent of f(x)=x^2+2,(-2,6) ?
The tangent of f(x)=x^2+2,(-2,6) is y=-4x-2