What is the tangent of g(x)=x^2-8,\at x=-3 ?

The tangent of g(x)=x^2-8,\at x=-3 is y=-6x-17

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tangent of g(x)=x^2-8,\at x=-3

tangent of

g (x) = x^{2} - 8, at x = - 3

y = - 6 x - 17

Solution steps

Find the tangent point:

(- 3, 1)

Find the slope of

g (x) = x^{2} - 8 : \frac{d g ( x )}{d x} = 2 x

Compute the slope of

g (x) = x^{2} - 8

x = - 3 : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 1) : y = - 6 x - 17

y = - 6 x - 17

\frac{d}{d x} (6 x^{2} + 9)

f (x) = \frac{( 2 x - 1 )}{( x + 5 )}, at x = 3

x \to 0 lim ((1 + x)^{\frac{4}{x}})

x^{2} y^{^{''}} + x y^{^{'}} + 4 y = 0

\frac{\partial}{\partial y} (3 x^{2} + y)

What is the tangent of g(x)=x^2-8,\at x=-3 ?
The tangent of g(x)=x^2-8,\at x=-3 is y=-6x-17