[graphing part 1][section 14]

(1.)  y = x2 + x - 30               here is the problem

0 = (x + 6)(x - 5)           factor and set equal to 0

x + 6 = 0    x - 5 = 0    set each factor equal to 0

-6 -6      +  5  +5   add this to each side
___________  ______________
x = -6  ,    x = 5         add

(ii.) y = 02 + 0 - 30    replace x with 0

y = -30               combine like terms

(iii.)   y' = 2x + 1     take the derivative

2x + 1 = 0   set the derivative equal to 0

-1  -1    subtract 1 from each side
_________________
2x  =  -1       subtract
____   ____
2     2       divide each side by 2

x = -1/2           cancel

y(-1/2) = (-1/2)2 + (-1/2) - 30    replace x with -1/2

y(-1/2) = -30.25            multiply and combine like terms

results:  (-6,0) and (5,0) are the x intercepts

(0, -30) is the y intercept

(-1/2, -30.25) is the relative minimum

(2.)  y = -x2 + 2x - 4

(i.)  y = -(0)2 + 2(0) - 4     replace x with 0

y = -4               multiply combine like terms

(ii.)   y' = -2x + 2     take the derivative

-2x + 2 = 0   set the derivative equal to 0

2x - 2 = 0     multiply thru by -1

+   2   +2   add 2 to each side
_______________
____   ____
2      2        divide each side by 2

x = 1        divide and cancel

y(1) = -(1)2 + 2(1) - 4    replace x with 1

y(1) = -3           multiply combine like terms

results:  (0, -4) is the y intercept

(1,-3) is the relative maximum point

(3.)   y= 4x2 - 8            here is the problem

y = 4(0)2 - 8       replace x with 0

y = -8       multiply combine like terms

(ii.)   4x2 - 8 = 0   set the function equal to 0

+8  +8   add 8 to each side
________________
____      ___
4         4         divide each side by 4

x2 = 2               divide and cancel

_          _
x =
2    x = -2        take square roots

(iii.)  y' = 8x           take the derivative

8x = 0   set the derivative equal to 0
___ ___
8     8        divide each side by 8

x = 0           divide and cancel

y(0) = 4(0)2 - 8     replace x with 0

y(0) = -8            multiply combine like terms

results:  (0,-8) is the y intercept
_           _
(
2,0) and (-2,0)  are the x - intercepts

(0,-8) is the relative minimum point

(4.)  y = (-3/2)x2 + 4x - 7     here is the problem

(i.)  y =  (-3/2)(0)2 + 4(0) - 7   replace x with 0

y = -7         multiply combine like terms

(ii.)  (-3/2)x2 + 4x - 7 = 0   set the function equal to 0

3x2 - 8x + 14 = 0     multiply thru by -2, cancel

b2 - 4ac             use the discriminant formula

=      (-8)2 - 4(3)(14)     make substitutions

=      64 - 168      multiply

=    -104              subtract

(iii.)   y' = -3x + 4    take the derivative

-3x + 4 = 0    set the derivative equal to 0

3x - 4 = 0      multiply thru by -1

+ 4  +4      add 4 to each side
_______________
____    ___
3       3          divide each side by 3

x = 4/3             cancel

y(4/3) = (-3/2)(4/3)2 + 4(4/3) - 7   replace x with 4/3

y(4/3) = (-3/2)(16/9) + (16/3) - 7      multiply

y(4/3) = (-48/18) + (16/3) - 7       multiply

y(4/3) = (-7/3) + (16/3) - 7         reduce

y(4/3) = -4                      combine like terms

results:  (0, -7) is the y intercept

There are no x intercepts

(4/3, -4) is the maximum point

(5.)   y = x + (1/x)          here is the problem

x = 0   is the vertical asymptote

y = x    is the oblique asymptote

y' = 1 - (1/x2)    take the derivative of y

1 - (1/x2) = 0     set the derivative equal to 0

+(1/x2)  +(1/x2)    add this to each side
____________________

x2 = 1       multiply each side by x2, cancel

x = 1   x = -1    take square roots

y = 1 + (1/1)     y = (-1) + (1/-1)     replace x with 1 & -1

y = 2              y = -2          divide and add

results:  (1,2) and (-1,-2) are extreme points

Here is the graph: (6.)  y = (x2 + 1)/(x + 1)        here is the problem

x = -1             this is the vertical asymptote

y = (02 + 1)/(0 + 1)    replace x with 0

y = 1                 add and divide

(0,1) is the y intercept

There is no x intercept.

-1 |    1    0     1   use synthetic division
___|        -1     1
_______________
1   -1     2

result:  x - 1 + [2/(x + 1)]

y = x - 1   is the oblique asymptote

y = (x2 + 1)/(x + 1)

y' = [2x(x + 1) - (x2 + 1)]/(x + 1)2  use the quotient rule

y' = (2x2 + 2x - x2 - 1)/(x + 1)2   multiply thru parentheses

y' = (x2 + 2x - 1)/(x + 1)2        combine like terms

x2 + 2x - 1 = 0    set the top equal to 0

+2  +2         add 2 to each side
________________
x2 + 2x + 1 = 2             add

(x + 1)2 = 2                  factor
_               _
x + 1 =
2     x + 1 = -2         take square roots

-1  -1         -1  -1     subtract 1 from each side
_____________ ____________
_               _
x = -1 +
2 ,   x = -1 - 2       subtract
_                _                              _
y = [(-1 +
2)2 + 1]/[(-1 + 2) + 1]   replace x with -1 + 2
_            _
y = (1 - 2
2 + 2 + 1)/(2)   square the binomial, add -1 & 1
_    _
y = (3 - 2
2)/(2)   combine like terms

_                               _
y = (3
2 - 4)/2      multiply thru by 2
_                _                               _
y = [(-1 -
2)2 + 1]/[(-1 - 2) + 1]    replace x wiht -1 - 2
_
y = [(1 + 2
2 + 2) + 1]/(-2)   square the binomial, add -1 & 1
_     _
y = (3 + 2
2)/(-2)   combine like terms
_                                   _
y = (-3
2 - 4)/(2)       multiply thru by -2

results: (0,1) is the y - intercept

There is no x intercept

y = x - 1   is the oblique asymptote
_     _                    _      _
(
-1 + √2, (3√2 - 4)/2) and (-1 - √2, (-3√2 - 4)/2)

are the extreme points.

(7.)  y = x3 + x             here is the problem

(i.)  y = (0)3 + (0)        replace x with 0

(0,0) is the y intercept

(ii.)  x3 + x = 0       set the function equal to 0

x(x2 + 1) = 0       factor

x = 0           set this factor equal to 0

(0,0) is the x intercept

(iii.)   y' = 3x2 + 1        take the derivative

3x2 + 1 = 0        set the derivative equal to 0

[no solution]

results:  (0,0) is the x intercept and the y intercept

and there are no extreme points

(8.)(i.)  y = x3 - 12x + 10              here is the problem

y = (0)3 - 12(0) + 10    replace x with 0

y = 10              multiply combine like terms

(0,10) is the y intercept

(ii.)   y' = 3x2 - 12         take the derivative

3x2 - 12 = 0    set the derivative equal to 0

+  12  +12    add 12 to each side
________________
____        ____
3           3         divide each side by 3

x2 = 4                divide and cancel

x = 2  and x = -2    take square roots

y(2) = (2)3 - 12(2) + 10   y(-2) = (-2)3 - 12(-2) + 10

[replace x with 2 and with -2]

y(2) = -6         y(-2) = 26   multiply combine like terms

results:  (0,10) is the y intercept

(2,-6) and (-2, 26) are the extremes

(9.)   y = x3 - x2      here is the problem

(0,0) is the y intercept

x3 - x2 = 0      set the function equal to 0

x2(x - 1) = 0      factor

x - 1 = 0      set this factor equal to 0

+ 1  +1   add 1 to each side
____________

(0,0) and (1,0) are the x intercepts

y' = 3x2 - 2x              take the derivative

3x2 - 2x = 0   set the derivative equal to 0

x(3x - 2) = 0         factor

x = 0    3x - 2 = 0  set each factor equal to 0

+  2  +2   add 2 to each side
______________
___    ___
3      3      divide each side by 3

x = 2/3            cancel

y = (2/3)3 - (2/3)2       replace x with 2/3

y = (8/27) - (4/9)              multiply

y = (8/27) - (12/27)      multiply 4/9 by 3/3

y = -4/27              combine like terms

results:  (0,0) and (2/3, -4/27) are the extremes

(10.)   y = x3 - 3x2 - 45x + 25

(0,25) is the y intercept

y' = 3x2 - 6x - 45         take the derivative

3x2 - 6x - 45 = 0   set the derivative equal to 0
___  ___  ___ ____
3     3    3    3     divide thru by 3

x2 - 2x - 15 = 0       divide

(x - 5)(x + 3) = 0        factor

x = 5 and x = -3

y(5) = (5)3 - 3(5)2 - 45(5) + 25

y(-3) = (-3)3 - 3(-3)2 - 45(-3) + 25

[replace x with 5][replace x with -3]

y(5) = 125 - 75 - 225 + 25        multiply

y(-3) = -27 - 27 + 135 + 25       multiply

y(5) = -150    y(-3) = 106    combine like terms

(5,-150) and (-3, 106) are the extreme points