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

Compute the slope of

y = 11 x^{2} - 4 x

x = a_{0} : 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}

y^{^{'}} = \frac{( sec ( y ) )}{( x - 4 ) ^{2}}

\frac{d y}{d x} = \frac{2 - e ^{x}}{3 + 2 y}, y (0) = 0

\int_{0}^{1} (x + 1)^{2} d x

\int_{0}^{π} tan^{2} (x) d x

\frac{d}{d x} (\frac{x ^{2} - x + 2}{x})

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