What is the tangent of-x^2+9x ?

The tangent of-x^2+9x is y=(-2a_{0}+9)x+a_{0}^2

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tangent of-x^2+9x

tangent of

- x^{2} + 9 x

y = (- 2 a_{0} + 9) 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} + 9 a_{0})

Find the slope of

y = - x^{2} + 9 x : \frac{d y}{d x} = - 2 x + 9

Compute the slope of

y = - x^{2} + 9 x

x = a_{0} : m = - 2 a_{0} + 9

Find the line with slope m=

- 2 a_{0} + 9

and passing through

(a_{0}, - a_{0}^{2} + 9 a_{0}) : y = (- 2 a_{0} + 9) x + a_{0}^{2}

y = (- 2 a_{0} + 9) x + a_{0}^{2}

\int 3 e^{- x} sin (x) d x

\frac{2 s - 4}{s ^{2} - 4 s + 13}

\frac{d}{d x} (- 0.5 x^{2} + 64 x - 25)

y^{^{''}} + y = 3 sin (x)

y^{^{''}} + y^{^{'}} = 6 sin (2 x)