CPhill

How's this???

Here's some for BOTH of YOU......

Now.....You two behave  !!!!

We're not always "boring" and "responsible"  Z-Man.....!!!!

y = 2(x + 1) (x +2) ......let's expand this to get this form ... y = ax^2 + bx + c

y = 2x^2 + 6x + 4  

The x coordinate of the vertex is given by  -b/2a  =  -6/(2*2)  = -6/4 =  -3/2  = -1.5

To find the y coordinate of the vertex, just plug this value into the function and we get

2(-1.5)^2 + 6(-1.5) + 4 = 

2(2.25) - 9  + 4 =

4.5 - 9 + 4 = -0.5

So the vertex is (-1.5, -0,5)

Yep....it is.....

I'll try (3), too

Here's the same figure as in the answer to (2)

Using reasoning from the other two answers, P cannot fall into area ABFE, because then, any triangle DPC is greater than any triangle APB.  P also cannot fall into area EID because all triangles BPC would be greater than all triangles APB.

If P falls into any of the regions IFC, ICG of IGD, all triangles APB will have greater areas than either triangles BPC or CPD. And these areas are 3/8 of the square ABCD = 37.5%.....so that's the probabilty that tiangle APB will have a greater area than either BPC or CPD.

log x + logx^2 + log x^3

Notice that the last two terms can be written as

2log x     and   3log x      so we have

log x + 2log x + 3logx  =

(1 + 2 + 3) log x  =

6 log x

Very, very nice MG1.......!!!!!

A BIG 3 points from me.....

I'll also take a stab at (2)

Consider square ABCD below:

This square is divided into 8 equal areas. It is clear that P cannot fall anywhere into rectangle ABFE because, although triangle APB and and triangle DPC would have equal bases, DPC would always have a greater height, and thus, a greater area.

It's also clear that P cannot fall into area IFC......To see this, note that if P fell on AC, its distance from AD and AB would be exactly the same. But, if P falls in area IFC, its distance from AD would always be greater than its distance from AB. Thus, triangles APB and APD formed by having P fall into this region would have equal bases, but APD would always have a greater height than triangle APB.

For the same reason, P cannot fall into area EID, because all triangles BPC would have greater heights than triangles APB.

So, P can only fall into region CID. Here triangles APB, BPC, CPD and DPA will all have the same bases, but APB  will have a height greater than the other three.

And since  CID is 25% of the area of the square, the probabilty that APB is greater than the other three mentioned triangles is ≈ 25%

I'll attempt to answer (1)

Shown below is square ABCD

Notice that if point P falls anywhere within rectangle EFCD, then the triangle APB will have a greater area than triangle DPC because each will have an equal base, but APB will have a greater height than DPC. And since rectangle EFCD is 1/2 of the area of the square, then the probability of P falling into this area is ≈ 1/2. So, the probability that APB > DPC   ≈ 1/2. [Of course, there is the slight possibility that P falls exactly on EF, which would make both triangles equal in area.......thus the need for the " ≈  "    symbol  !!!  ]