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

The tangent of f(x)=x^2-6x,\at x=-2 is y=-10x-4

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

tangent of

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

y = - 10 x - 4

Solution steps

Find the tangent point:

(- 2, 16)

Find the slope of

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

Compute the slope of

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

x = - 2 : m = - 10

Find the line with slope m=

- 10

and passing through

(- 2, 16) : y = - 10 x - 4

y = - 10 x - 4

\frac{d y}{d x} = \frac{( x ^{2} )}{y ( 1 + x ^{3} )}

\frac{d}{d x} (\frac{3}{( 1 - x ^{2} ) ^{4}})

\frac{d}{d x} (cos^{4} (x) - sin^{4} (x))

What is the tangent of f(x)=x^2-6x,\at x=-2 ?
The tangent of f(x)=x^2-6x,\at x=-2 is y=-10x-4