What is the tangent of f(x)=x^2-6,\at x=-5 ?

The tangent of f(x)=x^2-6,\at x=-5 is y=-10x-31

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tangent of f(x)=x^2-6,\at x=-5

tangent of

f (x) = x^{2} - 6, at x = - 5

y = - 10 x - 31

Solution steps

Find the tangent point:

(- 5, 19)

Find the slope of

f (x) = x^{2} - 6 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

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

x = - 5 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 19) : y = - 10 x - 31

y = - 10 x - 31

x \to 0 lim (\frac{3 ^{x} - 1}{x ^{2}})

\frac{d}{d t} (9 t)

\frac{\partial}{\partial x} (9 x e^{x y})

\frac{d}{d x} (arctan (\frac{3 x}{2}))

\int (tan (x))^{- 1} d x

What is the tangent of f(x)=x^2-6,\at x=-5 ?
The tangent of f(x)=x^2-6,\at x=-5 is y=-10x-31