What is the tangent of 2x^2+8x ?

The tangent of 2x^2+8x is y=(4a_{0}+8)x-2a_{0}^2

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tangent of 2x^2+8x

tangent of

2 x^{2} + 8 x

y = (4 a_{0} + 8) x - 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 8 a_{0})

Find the slope of

y = 2 x^{2} + 8 x : \frac{d y}{d x} = 4 x + 8

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 = 4 a_{0} + 8

Find the line with slope m=

4 a_{0} + 8

and passing through

(a_{0}, 2 a_{0}^{2} + 8 a_{0}) : y = (4 a_{0} + 8) x - 2 a_{0}^{2}

y = (4 a_{0} + 8) x - 2 a_{0}^{2}

What is the tangent of 2x^2+8x ?
The tangent of 2x^2+8x is y=(4a_{0}+8)x-2a_{0}^2