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

The tangent of f(x)=x^2+x,\at x=-4 is y=-7x-16

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

tangent of

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

y = - 7 x - 16

Solution steps

Find the tangent point:

(- 4, 12)

Find the slope of

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

Compute the slope of

f (x) = x^{2} + x

x = - 4 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 4, 12) : y = - 7 x - 16

y = - 7 x - 16

\int_{0}^{1} e^{x} d x

e^{x^{2}}

18 (x^{2} + 6)^{8}

\int \frac{x ^{2} - 7}{( x - 2 ) ( x + 1 )} d x

\int \frac{cot ^{2} ( x )}{sin ^{2} ( x )} d x

What is the tangent of f(x)=x^2+x,\at x=-4 ?
The tangent of f(x)=x^2+x,\at x=-4 is y=-7x-16