CPhill

x^4  + 8x^3  + 7x^2 - 40x  - 60  = 0 

Write as

 x^4 + 8x^3 +  [12x^2   -  5x^2]  - 40x - 60 ]  = 0

x^2 [x^2 + 8x + 12 ]  -  [ 5x^2 + 40x + 60 ]  = 0

x^2 [  (x + 6) (x + 2) ]  -  [ (5x + 10) (x + 6) ]  = 0

x^2 [ (x + 6) (x + 2) ] - [ 5 ( x + 2)  (x + 6) ]  = 0

[ (x + 2) (x + 6) ] [ x^2  - 5 ]  = 0

(x + 2) (x + 6) (x^2 - 5)  = 0

The rational  roots will come from setting the first two linear factors to 0 and solving for x

Thus.....   x  = -2    and x  = -6     are the rational roots

OK...OBT....see if this helps

Row 0  =  1 * 10^0      =  1  =   11^0

Row 1 =   1 * 10^1  + 1*10^0   =  11     =   11^1

Row 2 =   1*10^2  + 2*10^1 +  1 *10^0   =  121    =  11^2

Row 3 =   1*10^3  + 3*10^2 + 3*10 + 1*10^0  =  1331    =  11^3

Look  at row 5  which  is      1   5    10    10    5     1

So we have

1*10^5   +  5*10^4  + 10*10^3  + 10*10^2  + 5*10^1  + 1 * 10^0   = 161051  =  11^5

Do you see the pattern  ???

As x decreases without bound......the graph of f(x) increases without bound

As x increases without bound.......the graph of f(x) decreases  but is bounded below by 0

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The y intercept for the exponential function  is (0,10)

For the linear function......y decreases by 4  every time that x decreases by 1

So  when x decreases by 2, y decreases by 8

So.......from (2, 20)    the y intercept  is  ( 0, 12)

So    .... the linear function  is  2 more

8x⁴ / 2y² z Multiplied by 5yz² / 12x³ z

Multipy the numerators

( 8x⁴) (5yz²)  =  8*5x⁴yz²  =  40x⁴yz²       (1)

Multiply the denominators

(2y²z) (12x³z)  =  2*12  * y² * x³ * z * z   =   24y²x³z²    (2)

Put   (1)  over (2)

40x⁴yz²                 40        x⁴      y       z²                     5x

________     =      __       __     __     ___     =        ___  

 24y²x³z²              24       x³      y²      z²                     3y

Not an expert here....but notice that, working backwards

11⁰   =  1 

11¹  =  (10 + 1)¹  =  C(1,0)* 10  + C(1,1) * 1  =  10  + 1 =  11

11²  = 

(10 + 1)²  =  C(2, 0) * 10²  + C(2, 1) * 10  + C(2, 2) * 1   = 

100 + 2*10 + 1  =

121

11³ =

(10 + 1)³   = 

C(3, 0) 10³  + C(3,1)* 10²  + C(3,2) * 10 + C(3,3) * 1   =

10³    + 3 * 10²  +  3 * 10   +  1  =

1000  +  300  +  30  +  1  =

1331

etc.........

OK.....here you go....!!!!

4 *2^{x - 2}    

The points that appear are

4 *2 ^{5- 2}   =  4 * 2³   =  4 * 8   =  32    ⇒  ( 5, 32)

4 *2 ^{2- 2}   =  4 * 2⁰   =  4 * 1   =  4    ⇒  ( 2, 4)    

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3(1/2)^x

The points that appear are

3(1/2)⁰  = 3 * 1   = 3  ⇒  (0, 3)

None of them

Here is the graph  :  https://www.desmos.com/calculator/6t93qupx9m

Are you sure that you have the correct function ???

f)x)  =  (-5)^{x + 3}

The points that appear are

(-5)^{0 + 3}   =  (-5)³  =  -125   ⇒  (0, -125)

(-5) ^{-5+ 3}   =  (-5)^-2  =  1/5²    =    1 /25 ⇒  (-5, 1/25)

Based on this......can you figure out the other two ??

If not......let me know......

The "12x"   part is known as the variable part

The  "24"   part is known as the fixed part....it is the amount of elecricity that is used independent of the  additional electricity used in the manufacturing process...so.. even if no parts are manufactured, 24 watts are still used 

Note  that the time to double the money is independent of the amount borrowed.....thus.... it takes as long for $1  to double as it does for $1000 to double  [ believe it, or not  !!!]

So  calling the original amount, A..... we have that

2A  = A (1.05)^N      where N is in months.....divide both sides by A

2 = (1.05)^N        take the log of both sides

log 2   =  log (1.05)^N     and by a log property, we can write

log 2  =  N * log (1.05)      divide both sides by log (1.05)

log (2)  / log (1.05) =  N   ≈  14.2  months

And the least integer greater or equal to  14.2  = 15