What is the tangent of y=x^2+4x+9,\at x=8 ?

The tangent of y=x^2+4x+9,\at x=8 is y=20x-55

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tangent of y=x^2+4x+9,\at x=8

tangent of

y = x^{2} + 4 x + 9, at x = 8

y = 20 x - 55

Solution steps

Find the tangent point:

(8, 105)

Find the slope of

y = x^{2} + 4 x + 9 : \frac{d y}{d x} = 2 x + 4

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 = 20

Find the line with slope m=

20

and passing through

(8, 105) : y = 20 x - 55

y = 20 x - 55

\frac{d}{d x} (x^{2} - 4)

\int_{0}^{1} 2 e^{x} d x

x \to 0 - lim (\frac{2}{x})

x^{^{'}} = x - 2

\frac{d y}{d t} = \frac{1}{( 2 y + 1 )}

What is the tangent of y=x^2+4x+9,\at x=8 ?
The tangent of y=x^2+4x+9,\at x=8 is y=20x-55