Question

Find p and q, if Definition area is [3;10]

Answer 1

$\\ \tt\longmapsto y=\sqrt{-x^2+px+q}$

• Put(3,10)

$\\ \tt\longmapsto y^2=-x^2+px+q$

$\\ \tt\longmapsto 10^2=-(3^)2+3p+q$

$\\ \tt\longmapsto 100=-9+3p+q$

$\\ \tt\longmapsto 3p+q=109$

• Use hit and trial method.

Let p,q be (27,28)

Apply

$\\ \tt\longmapsto 3(27)+28=81+28=109$

Verified