What is the tangent of y=3x^2+10x+2,\at x=1 ?

The tangent of y=3x^2+10x+2,\at x=1 is y=16x-1

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tangent of y=3x^2+10x+2,\at x=1

tangent of

y = 3 x^{2} + 10 x + 2, at x = 1

y = 16 x - 1

Solution steps

Find the tangent point:

(1, 15)

Find the slope of

y = 3 x^{2} + 10 x + 2 : \frac{d y}{d x} = 6 x + 10

Compute the slope of

y = 3 x^{2} + 10 x + 2

x = 1 : m = 16

Find the line with slope m=

16

and passing through

(1, 15) : y = 16 x - 1

y = 16 x - 1

x \to \infty lim (2^{2 x})

y = \frac{3 x - 1}{6 x + 5}

(\frac{1}{2} arcsin (x) + \frac{1}{4} sin (2 arcsin (x)))^{^{'}}

x \cdot g (x) \cdot f (x)

f (x) = (6 x^{2} - x + 7)^{- \frac{5}{3}}

What is the tangent of y=3x^2+10x+2,\at x=1 ?
The tangent of y=3x^2+10x+2,\at x=1 is y=16x-1