f(10−x)=3x2+4x−5 and f(x)=px2+qx+r Let 10−x=y, then x=10−y f(y)=3(10−y)2+4(10−y)−5 =3(100+y2−20y)+40−4y−5 =340+3y2−60y−4y−5 =3y2−64y+335 Replace y by x⇒f(x)=3x2−64x+335 compare with f(x)=px2+qx+r, we obtain p=3,q=−64,r=335 ⇒p+q+r=274