CPhill

AB  is a horizontal line.....the slope of a horizontal line   = 0

Its equation  is      y  = -2

 3(x - 4) + 2(x^2 - x + 7) - 5(x - 1)   

Constant is

3 * (- 4 )  +  2 * 7  + (-5)* (-1)  =

-12   +   14   +    5    =

7   ......just as you found ....good job....!!!

Let's first find  g (g(x)) 

This is literally putting g(x)  into the x in  g(x)....so we have.... 

2 [ 2x - 1]  -  1    =    4x - 3

So  

g (g(x))  - g (x)   =

[ 4x - 3 ] - [ 2x  - 1 ]  =

2x  - 2    =  f(x)

So

f(3)   =  2(3) - 2     =   4

Triangle  PQR   is similar to triangle ABC

So

PQ / QR   =   AB / BC

PQ / 40  =  21 / 35      multiply both sides by  40

PQ  =  (40/35) * 21  =  (8/7) * 21  = 24  =  "a"

And

PR / PQ  =  AC / AB

PR / 24  =  28 / 21   multiply both sides by 24

PR =  24 (28/21)  =  24 (4/3)  = 32  =  "b"

(7+8x)-3(2+6x+2x^2)+9(1+3x+4x^2+12x^3)-9(7-x^2-8x^3+13x^4)

8x  -  18x +  27x   =

17x

So.....the coefficient is  17 

Very nice detective work, Melody!!!

I haven't seen this done before, but it appears to work!!!

Since the polynomial is symmetric about y and appears to intercept the x axis twice, I would have guessed that it was a quadratic......but yours perfectly models the data....

Good job, Manuel   !!!!

Multiplying each by a constant does nothing to the degree

The degree  of these two added polynomials will be the degree of the highest one..i.e., 5th degree

(a, b)   (a + 3, b+ 3)

The slope is     [  b + 3 - b ]  / [ a + 3 - a ]   =  3 / 3   =  1

Point-slope form

y - b  =  1 ( x - a)

Slope-intercept form

y  = x - a + b

Or

y =  x + (b - a)

1 1/5   /    3

1  1/5  =  6/5  

3  =  15/5     

So we have

 6

 5                                                                    

__         =  "cancel" the denominators          6    =      2

                                                                    15           5

15

 5