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

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

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

tangent of

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

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

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} + 3 a_{0} + 3)

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

- 3 a_{0} + 3

and passing through

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

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

\int \frac{1}{r} d r

\frac{d}{d x} (\frac{x ^{2}}{9 + 2 x})

x^{2} + 2 x y - y^{2} + x = 12, (3, 6)

\frac{3}{s} + 8 \frac{1}{s ( s + 1 )}

\int sin (x) - 2 cos (x) d x

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