What is the tangent of f(x)=x^2+9x,\at x=-8 ?

The tangent of f(x)=x^2+9x,\at x=-8 is y=-7x-64

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tangent of f(x)=x^2+9x,\at x=-8

tangent of

f (x) = x^{2} + 9 x, at x = - 8

y = - 7 x - 64

Solution steps

Find the tangent point:

(- 8, - 8)

Find the slope of

f (x) = x^{2} + 9 x : \frac{df ( x )}{d x} = 2 x + 9

Compute the slope of

f (x) = x^{2} + 9 x

x = - 8 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 8, - 8) : y = - 7 x - 64

y = - 7 x - 64

\int x (ln (x) + e^{x}) d x

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

\int \frac{21}{25 + 4 r ^{2}} d r

\frac{1}{x} - 1

\frac{\partial}{\partial x} (2 y - 3 x + 5)

What is the tangent of f(x)=x^2+9x,\at x=-8 ?
The tangent of f(x)=x^2+9x,\at x=-8 is y=-7x-64