What is the tangent of f(x)=-x^2-4x+1,\at x=-4 ?

The tangent of f(x)=-x^2-4x+1,\at x=-4 is y=4x+17

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tangent of f(x)=-x^2-4x+1,\at x=-4

tangent of

f (x) = - x^{2} - 4 x + 1, at x = - 4

y = 4 x + 17

Solution steps

Find the tangent point:

(- 4, 1)

Find the slope of

f (x) = - x^{2} - 4 x + 1 : \frac{df ( x )}{d x} = - 2 x - 4

Compute the slope of

f (x) = - x^{2} - 4 x + 1

x = - 4 : m = 4

Find the line with slope m=

4

and passing through

(- 4, 1) : y = 4 x + 17

y = 4 x + 17

f (x) = ln (8 x)

\int 8 (x + 1) d x

\int \frac{( x ^{5} - 2 x ^{2} )}{( 1 + x ^{3} ) ^{\frac{3}{2}}} d x

f (x) = \frac{1}{x + 1}

What is the tangent of f(x)=-x^2-4x+1,\at x=-4 ?
The tangent of f(x)=-x^2-4x+1,\at x=-4 is y=4x+17