CPhill

 x ^3 +4x ^2+7x+6

-2 is a root because   (-2)^3  + 4(-2)^2  + 7(-2)  + 6  =   -8 + 16  - 14 + 6  = 0

So  (x + 2)  is a factor.....using synthetic division, we can find the remaining polynomial as follows

-2  [ 1    4     7      6  ]

            -2     -4    -6

     _______________

       1    2     3      0

The remaining polynomial is

x^2  + 2x  + 3   set to 0

x^2 + 2x + 3   = 0        subtract 3 from both sides

x^2  + 2x    =  -3        take 1/2 of 2  = 1...square it  = 1...add it to both sides

x^2 + 2x  + 1   = -3 + 1      factor the left side, simplify the right

(x + 1)^2   = -2       take both roots

x + 1  = ±√[-2]

x = 1  = ±√[2] i     subtract 1 from both sides

x  =  ±√[2] i  - 1

So...the other two factors are    -1 + √[2] i  and   -1 - √[2] i

So...the complete factorization is

A. f(x)=(x+2)(x−(−1+i√2))(x−(−1−i√2))

x^3−4x^2+4x−16      factor by grouping

x^2 ( x - 4)  + 4 ( x - 4)   =

(x - 4) (x^2 + 4)   =

(x - 4)  ( x + 2i) ( x - 2i)

By the Remainder Theorem.....to find the remainder...just plug 3 into the function and we have

4(3)^3  - 17(3)^2  + 14(3)  - 3 = 

4*27  - 17*9  + 42  - 3  =

-6

Note that synthetic division would provide the same result

3 [  4   -17     14     -3   ]

            12    -15     -3

     ________________

      4    -5     -1      -6

Let ABC be an isosceles triangle  with  AB  = AC

Draw median BM to side  AC  and  median CN to side AB

Note that  since AB  = AC...then then the medians divide  AB and AC into two equal parts.....so....MC   = NB

And since  AB  = AC, then angle ABC  = angle ACB

And BC  = BC

So...by side-angle side.....triangle  MCB is congruent to triangle NBC

So...therefore....BM  = CN 

See this image : 

OK...look at the graph....

Note that the graph is  above  y =  0  from  x  =   -infinity to x =  - 1   [but not including -1 ]

And the graph  is equal to  or above  y = 0  from x = 0   0 to x  =  1     [ icluding 0, but not including 1]

Ad the graph is equal to or above  y  = 0  from x =   2  to  x =  infinity  [  including 2 ]

My notation means the same thing

x ≥ 2  is the same as  [2,  infinity ) 

x < -1   is the same  as   ( - infinity, -1)

 0 ≤ 𝑥 < 1  is the same as  [0, -1 )

y  = [ x ( x - 2) ] / [ x^2 - 1 ] =    [x^2 - 2x] / [ (x + 1) ( x - 1) ]   = (x^2 - 2x) / [ x^2 - 1)

a)  Setting both linear factors in the denominator to 0 and solving for x gives us the vertical asyptotes of  x  = -1 and x  = 1

Because the coefficients on each of the x^2 terms in the numerator/ denominator  are 1, their ratio  gives the  horizontal asymptote of  y  = 1 / 1    =  1

b) To see where it crosses the  x axis, let y  = 0  and we have

0  = x (x - 2)  / (x^2 - 1)     multiply both sides by  x^2 - 1

0 =  x ( x - 2)   set both factors to  0  and solve for x and we get the x intercepts of x = 0  and  x  = 2

To see where it crosses the y  axis, let x  = 0   and we have  0 (0 - 2) / ( 0^2 - 1)   =  0

c) Here is the curve : https://www.desmos.com/calculator/i0vbehvizi

d) Inspection of the graph reveals that the intervals that solve this inequality  are  (-infinity ,-1) U [0, 1) U [2, infinity )

(x^3 - 2)  / (x^2 + x) 

Partial faction decomposition can only be done whenever the degree of the polynomial in the numerator  is <  the degree of the polynomial in the denominator

Performing polynomial division, we have

                 x        - 1

x^2 + x  [  x^2   +  0x^2 + 0x   - 2 ]  

                x^3   + x^2

               ___________________

                          -x^2  + 0x  

                          -x^2  - 1x

                         _________

                                      x    -  2

The result  is   ( x - 1)  +  ( x - 2) / ( x^2 + x)

We can decompose the  remainder  as

( x - 2) / [ x ( x + 1)   =   A / x   +   B / ( x + 1)

Multiply through by    x ( x + 1)

x -  2  =  A(x + 1)  + Bx   simplify

x - 2 =  (A + B)x  + A       equate coefficients

A + B  = 1

A  =  -2

Which means that B  = 3

So...the complete  decomposition is

 (x -1 )  - [ 2 / x ]   +  [ 3 / ( x + 1) ] 

Second one

a)

Linear regression equation         Coefficient of determination  (r^2)

y = 2.16x + .391                         .903

Quadratic regression equation   Coefficient of determination

y = .096x^2 + .714x  + 3.803             .948

Exponential regression equation     Coefficient of determination

y = 4.63 (1.152)^x                               .951

b)  The exponetial regression equation fits the data best   [ highest r^2   and r value  ]

First one....

(a)   The two points that should be used are ≈  (1,14)  and ≈  (7,7)....these don't lie directly on the line, but they are very close to the line

(b)  Slope =  [ 7 - 14 ] / [ 7 - 1 ] =   -7 / 6  ≈  - 1.166

(c)  Equation of the linear model  in point- slope form  is

(y - 7) = -1.166(x - 7)

(d)  Equation of the linear model in slope-intercept form :

y  = -1.166x + 8.162 + 7

y = -1.166x + 15.162