What is the tangent of y=x^2+2x+10,\at x=6 ?

The tangent of y=x^2+2x+10,\at x=6 is y=14x-26

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tangent of y=x^2+2x+10,\at x=6

tangent of

y = x^{2} + 2 x + 10, at x = 6

y = 14 x - 26

Solution steps

Find the tangent point:

(6, 58)

Find the slope of

y = x^{2} + 2 x + 10 : \frac{d y}{d x} = 2 x + 2

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 = 14

Find the line with slope m=

14

and passing through

(6, 58) : y = 14 x - 26

y = 14 x - 26

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

x \to 0 lim (2400 x - 2 x^{2})

x \to 1 lim (\frac{x}{( x - 1 )})

y = 6 x^{3} - x^{2}

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 6, at x = 0

What is the tangent of y=x^2+2x+10,\at x=6 ?
The tangent of y=x^2+2x+10,\at x=6 is y=14x-26