What is the tangent of-x^2+6x ?

The tangent of-x^2+6x is y=(-2a_{0}+6)x+a_{0}^2

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tangent of-x^2+6x

tangent of

- x^{2} + 6 x

y = (- 2 a_{0} + 6) 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} + 6 a_{0})

Find the slope of

y = - x^{2} + 6 x : \frac{d y}{d x} = - 2 x + 6

Compute the slope of

y = - x^{2} + 6 x

x = a_{0} : m = - 2 a_{0} + 6

Find the line with slope m=

- 2 a_{0} + 6

and passing through

(a_{0}, - a_{0}^{2} + 6 a_{0}) : y = (- 2 a_{0} + 6) x + a_{0}^{2}

y = (- 2 a_{0} + 6) x + a_{0}^{2}

\frac{d}{d x} (\frac{8}{1 + x})

f (x) = ln (tan (x))

\frac{d x}{d t} = \frac{6 x ^{2} + t t ^{2} + x ^{2}}{6 t x}

\int_{0}^{3} (- x^{3} + 3 x^{2} - 2) d x

\int (\frac{3}{x ^{5}}) d x