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

The tangent of f(x)=-x^2-4x-6,\at x=4 is y=-12x+10

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tangent of f(x)=-x^2-4x-6,\at x=4

tangent of

f (x) = - x^{2} - 4 x - 6, at x = 4

y = - 12 x + 10

Solution steps

Find the tangent point:

(4, - 38)

Find the slope of

f (x) = - x^{2} - 4 x - 6 : \frac{df ( x )}{d x} = - 2 x - 4

Compute the slope of

f (x) = - x^{2} - 4 x - 6

x = 4 : m = - 12

Find the line with slope m=

- 12

and passing through

(4, - 38) : y = - 12 x + 10

y = - 12 x + 10

\int cos (\frac{π t}{2}) d t

\int \frac{tan ^{3} ( ln ( x ) )}{8 x} d x

\frac{d ^{2}}{d x ^{2}} y = sin (x)

\int (2 x^{3}) d x

What is the tangent of f(x)=-x^2-4x-6,\at x=4 ?
The tangent of f(x)=-x^2-4x-6,\at x=4 is y=-12x+10