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

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 = 16

Find the line with slope m=

16

and passing through

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

y = 16 x - 1

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