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

The tangent of f(x)=x^2+x,\at x=-4,-1 is y=-7x-16

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

tangent of

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

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 - \frac{2}{y} d y

\int_{2}^{5} \frac{4}{9 + ( x - 2 ) ^{2}} d x

f (x) = ln (169 sin^{2} (x))

\int e^{t a n (8 x)} sec^{2} (8 x) d x

\frac{d}{d x} (8 x + 6 x^{\frac{3}{4}})

What is the tangent of f(x)=x^2+x,\at x=-4,-1 ?
The tangent of f(x)=x^2+x,\at x=-4,-1 is y=-7x-16