What is the tangent of f(x)=-3/2 x^2+3x+3 ?

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

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tangent of f(x)=-3/2 x^2+3x+3

tangent of

f (x) = - \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

f (x) = - \frac{3}{2} x^{2} + 3 x + 3 : \frac{df ( x )}{d x} = - 3 x + 3

Compute the slope of

f (x) = - \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

(e^{x} - 2)^{\frac{1}{2}}

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

\int sin (9 x) cos (3 x) d x

x \to 7 + lim (7 x - 7)

\int (4 x - 9) d x

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