What is the tangent of-18x^2+9x+30 ?

The tangent of-18x^2+9x+30 is y=(-36a_{0}+9)x+18a_{0}^2+30

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tangent of-18x^2+9x+30

tangent of

- 18 x^{2} + 9 x + 30

y = (- 36 a_{0} + 9) x + 18 a_{0}^{2} + 30

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 18 a_{0}^{2} + 9 a_{0} + 30)

Find the slope of

y = - 18 x^{2} + 9 x + 30 : \frac{d y}{d x} = - 36 x + 9

Compute the slope of

y = - 18 x^{2} + 9 x + 30

x = a_{0} : m = - 36 a_{0} + 9

Find the line with slope m=

- 36 a_{0} + 9

and passing through

(a_{0}, - 18 a_{0}^{2} + 9 a_{0} + 30) : y = (- 36 a_{0} + 9) x + 18 a_{0}^{2} + 30

y = (- 36 a_{0} + 9) x + 18 a_{0}^{2} + 30

x \to 0 - lim (x^{\frac{2}{3}})

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

(p^{3} + 1)^{- 1}

\frac{d}{d x} (cos (ln (x^{2})))

2 cos (3 x)

What is the tangent of-18x^2+9x+30 ?
The tangent of-18x^2+9x+30 is y=(-36a_{0}+9)x+18a_{0}^2+30