What is the tangent of y=(10x^2-6x+3)(1+2x),\at x=1 ?

The tangent of y=(10x^2-6x+3)(1+2x),\at x=1 is y=56x-35

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tangent of y=(10x^2-6x+3)(1+2x),\at x=1

tangent of

y = (10 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

y = 56 x - 35

Solution steps

Find the tangent point:

(1, 21)

Find the slope of

y = (10 x^{2} - 6 x + 3) (1 + 2 x) : \frac{d y}{d x} = 60 x^{2} - 4 x

Compute the slope of

y = (10 x^{2} - 6 x + 3) (1 + 2 x)

x = 1 : m = 56

Find the line with slope m=

56

and passing through

(1, 21) : y = 56 x - 35

y = 56 x - 35

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What is the tangent of y=(10x^2-6x+3)(1+2x),\at x=1 ?
The tangent of y=(10x^2-6x+3)(1+2x),\at x=1 is y=56x-35