Let `f(x)=a x^2+b x+c` be a quadratic expression having its vertex at (3, -2) and value of `f(0)=10. F in df(x)dot`

Explanation

`f(x) = ax^(2) + bx + c`
Vertex `equiv (-(b)/(2a),-(D)/(4a))`
`rArr -(b)/(2a) = 3` (1)
and ` -((b^(2) - 4ac))/(4a) = -2` (2)
From (2), we get
`-(b^(2) )/(4a) + 10 = -2 [ because c = f (0) = 10 ]`
or `-(b)/(2a)xx(b)/(2) = -12`
or `(3b)/(2) = -12`
or ` b = - 8`
`rArr a = (4)/(3)`
`rArr f(x) = (4)/(3) x^(2) - 8 x + 10`