What is the tangent of 6x^2+5x+8,\at x=4 ?

The tangent of 6x^2+5x+8,\at x=4 is y=53x-88

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tangent of 6x^2+5x+8,\at x=4

tangent of

6 x^{2} + 5 x + 8, at x = 4

y = 53 x - 88

Solution steps

Find the tangent point:

(4, 124)

Find the slope of

y = 6 x^{2} + 5 x + 8 : \frac{d y}{d x} = 12 x + 5

Compute the slope of

y = 6 x^{2} + 5 x + 8

x = 4 : m = 53

Find the line with slope m=

53

and passing through

(4, 124) : y = 53 x - 88

y = 53 x - 88

x \to 4 lim (x^{3} + 6 x)

x \to 2 lim (3 cos (x))

f (x) = x^{2} cos (\frac{1}{x})

(1 - 2 x)^{- 2}

n = 2 \sum \infty \frac{n}{3 n ^{3} + 1}

What is the tangent of 6x^2+5x+8,\at x=4 ?
The tangent of 6x^2+5x+8,\at x=4 is y=53x-88