What is the tangent of-3/2 x^2+9x+9 ?

The tangent of-3/2 x^2+9x+9 is y=(-3a_{0}+9)x+3/2 a_{0}^2+9

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tangent of-3/2 x^2+9x+9

tangent of

- \frac{3}{2} x^{2} + 9 x + 9

y = (- 3 a_{0} + 9) x + \frac{3}{2} a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{3}{2} a_{0}^{2} + 9 a_{0} + 9)

Find the slope of

y = - \frac{3}{2} x^{2} + 9 x + 9 : \frac{d y}{d x} = - 3 x + 9

Compute the slope of

y = - \frac{3}{2} x^{2} + 9 x + 9

x = a_{0} : m = - 3 a_{0} + 9

Find the line with slope m=

- 3 a_{0} + 9

and passing through

(a_{0}, - \frac{3}{2} a_{0}^{2} + 9 a_{0} + 9) : y = (- 3 a_{0} + 9) x + \frac{3}{2} a_{0}^{2} + 9

y = (- 3 a_{0} + 9) x + \frac{3}{2} a_{0}^{2} + 9

\int (12 x^{3} - 4 x - 2) d x

\int \frac{1}{\frac{1}{v}} d v

\frac{e ^{- 2 s}}{s ^{2} + 1}

5 x^{2} + y^{2} = 9

\int x^{3} x + 1 d x

What is the tangent of-3/2 x^2+9x+9 ?
The tangent of-3/2 x^2+9x+9 is y=(-3a_{0}+9)x+3/2 a_{0}^2+9