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

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

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tangent of-x^2+4x

tangent of

- x^{2} + 4 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = - x^{2} + 4 x

x = a_{0} : m = - 2 a_{0} + 4

Find the line with slope m=

- 2 a_{0} + 4

and passing through

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

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

\int_{0}^{\frac{π}{3}} 4 sec^{3} (x) d x

\int \frac{2}{3} y e^{y} d y

\frac{d}{d x} (- 2 x^{3} - 7 x + 5)

\frac{\partial}{\partial x} (y^{0.25} x^{0.75})

f (x) = x^{π}