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

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

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tangent of y=x^2+9x,\at x=-8

tangent of

y = x^{2} + 9 x, at x = - 8

y = - 7 x - 64

Solution steps

Find the tangent point:

(- 8, - 8)

Find the slope of

y = x^{2} + 9 x : \frac{d y}{d x} = 2 x + 9

Compute the slope of

y = 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

\frac{d}{d x} (- cos (e^{x}))

\int_{1}^{6} \frac{1}{2 x + 3 + x} d x

\frac{d}{d x} (7 x (x^{3} - 8 x + 2))

x \frac{d y}{d x} - 2 y = 5 x^{3}

\frac{\partial}{\partial x} (\frac{x ^{2}}{sin ^{2} ( x )})

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