What is the tangent of-6x^2+5x ?

The tangent of-6x^2+5x is y=(-12a_{0}+5)x+6a_{0}^2

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tangent of-6x^2+5x

tangent of

- 6 x^{2} + 5 x

y = (- 12 a_{0} + 5) x + 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 6 a_{0}^{2} + 5 a_{0})

Find the slope of

y = - 6 x^{2} + 5 x : \frac{d y}{d x} = - 12 x + 5

Compute the slope of

y = - 6 x^{2} + 5 x

x = a_{0} : m = - 12 a_{0} + 5

Find the line with slope m=

- 12 a_{0} + 5

and passing through

(a_{0}, - 6 a_{0}^{2} + 5 a_{0}) : y = (- 12 a_{0} + 5) x + 6 a_{0}^{2}

y = (- 12 a_{0} + 5) x + 6 a_{0}^{2}

y = (ln (x))^{3}

\int \frac{x + 3}{x - 1} d x

f (x) = e^{2 - x}

\frac{\partial}{\partial x} (e^{- x} tan (y))

\int + \frac{54}{1 - cos ( 9 x )} d x

What is the tangent of-6x^2+5x ?
The tangent of-6x^2+5x is y=(-12a_{0}+5)x+6a_{0}^2