What is the tangent of y=x^2+5x,(-4,-4) ?

The tangent of y=x^2+5x,(-4,-4) is y=-3x-16

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tangent of y=x^2+5x,(-4,-4)

tangent of

y = x^{2} + 5 x, (- 4, - 4)

y = - 3 x - 16

Solution steps

Find the slope of

y = x^{2} + 5 x : \frac{d y}{d x} = 2 x + 5

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 3

Find the line with slope m=

- 3

and passing through

(- 4, - 4) : y = - 3 x - 16

y = - 3 x - 16

\frac{d}{d x} (x - \frac{x ^{2}}{800})

\frac{d}{d x} (x^{3} - 6 x + 12)

\frac{d}{d x} (- 216 cos (3 x))

\int_{0}^{1} (1 - e^{- t})^{2} d t

\int - 2 x sin (x^{2}) d x

What is the tangent of y=x^2+5x,(-4,-4) ?
The tangent of y=x^2+5x,(-4,-4) is y=-3x-16