What is the tangent of y=x^2+3.5x+4pi,\at x=-4 ?

The tangent of y=x^2+3.5x+4pi,\at x=-4 is y=-4.5x+4pi-16

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tangent of y=x^2+3.5x+4pi,\at x=-4

tangent of

y = x^{2} + 3.5 x + 4 π, at x = - 4

y = - 4.5 x + 4 π - 16

Solution steps

Find the tangent point:

(- 4, 4 π + 2)

Find the slope of

y = x^{2} + 3.5 x + 4 π : \frac{d y}{d x} = 2 x + 3.5

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 = - 4.5

Find the line with slope m=

- 4.5

and passing through

(- 4, 4 π + 2) : y = - 4.5 x + 4 π - 16

y = - 4.5 x + 4 π - 16

x \to 0 + lim (\frac{x ^{2} e ^{\frac{1}{x}}}{1 + x ^{2}})

y^{^{'}} = \frac{x + 3}{y + 8}

f (x) = x^{2} - 9 x + 118

\frac{d}{d x} (x^{\frac{- 1}{2}} sin (x))

\int 3 (3 x + 2) d x

What is the tangent of y=x^2+3.5x+4pi,\at x=-4 ?
The tangent of y=x^2+3.5x+4pi,\at x=-4 is y=-4.5x+4pi-16