Let the analytical expression of the function be y=kx.
Because it passes point (2, -3a), it is: -3a=k*2, and a=-2k/3 (1) is obtained.
If it passes through point (a, -6), then: -6=k*a (2)
Substituting (1) into (2) gives -6 = k *(2k/3).
Finishing: k? =9 k= 3
And because the straight line passes through the fourth quadrant, k < 0 means k=-3.
Therefore, the analytical expression of this function is y=-3x.
Question 9
1) If P is the intersection of PC⊥X axis and C is the intersection of X axis, then PC = YOC = X.
s = 1/2 * OA * PC = 1/2 * 6 * y = 3y
X+y=8 so: S=3*(8-x)=24-3x.
24-3x > 0 0 < x < 8, because s > 0.
2) When the abscissa of point P is 5, that is, x=5.
△△OPA area S=24-3*5=9
3) The area of△ OPA should not be greater than 24.
Because the area of △OPA is s = 24-3x0 < x < 8.
So the area s is 0.