CPhill

t^2 + 2t - 24

__________

t^2  -  36

(  t + 6) ( t - 4)

____________

(t + 6) ( t  - 6)

(t - 4)

____          t  not equal to -6 or 6

(t - 6)

7) 

     x - 6                               x  -  6

   ___________   =         __________

  x^2 + 5x + 6                  (x + 2) ( x + 3)

We have no "holes" (one of the factors would have to cancel for this to happen )....the vertical asymptotes are at x = -3  and x = -2

8)  We have   a     same degree / same degree      situation....thus....the horizontal asymptote will be the ratio of the lead coefficients in each polynomial  =  -3/9   =  -1/3

Here's a pic

We have two areas.....one is a parallelogram  with a base of 6  and a height of  3

So.......its area  is :   base * height  =  6 * 3   =  18 units^2

We also have a triangle with the same base and a height of 2....so its area is :

(1/2) base * height  =  (1/2)(6) (2)  =  6 units^2

So.....the total area is   (18 + 6)   =  24 units^2

If  we have a root  at  (-3,0)  with a multiplicity of 2  and the point (2,0) is aslo on the graph.....we have three roots.....but this is a polynomial of degree 2, so it can't possibly have three roots.....did you make a mistake ??

Well....graphing is the fastest....however....the solution points could be difficult to interpret in some cases

But.....without a calculator or grapher, substitution is the fastest

(f/ g) (x)   =

   4x^2 + 6x

_____________

 2x^2 + 13x + 15

   2x(x + 6)

____________

(2x + 3) (x + 5)

To find the height, H, we can use :

sin (45°) =  H  / 80      rearrange as

H  =  80 *  sin (45°)   ≈  56.6  ft  

There isn't a 'cookbook" metrhod  for these.....your approach (as long as it's sound) is as good as any.....!!!

Very nice, X²   !!!

an  = 2(1/3)^{(n - 1)}