What is the tangent of 11x^2-4x ?

The tangent of 11x^2-4x is y=(22a_{0}-4)x-11a_{0}^2

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tangent of 11x^2-4x

tangent of

11 x^{2} - 4 x

y = (22 a_{0} - 4) x - 11 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 11 a_{0}^{2} - 4 a_{0})

Find the slope of

y = 11 x^{2} - 4 x : \frac{d y}{d x} = 22 x - 4

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 = 22 a_{0} - 4

Find the line with slope m=

22 a_{0} - 4

and passing through

(a_{0}, 11 a_{0}^{2} - 4 a_{0}) : y = (22 a_{0} - 4) x - 11 a_{0}^{2}

y = (22 a_{0} - 4) x - 11 a_{0}^{2}

a re a y = 2 x^{2}, y = 16 x - 32, y = 0

\frac{d y}{d t} = \frac{y + 1}{t + 1}

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

\int t ln (t + 2) d t

\frac{d}{d t} (sin (5 t))

What is the tangent of 11x^2-4x ?
The tangent of 11x^2-4x is y=(22a_{0}-4)x-11a_{0}^2