Intermediate Algebra: A Graphing Approach

1
Intermediate Algebra A Graphing Approach

• 8.3 Solving Equations by Using Quadratic
  Methods
• 8.4 Nonlinear Inequalities in One Variable
• 8.5 Quadratic Functions Their Graphs
• 8.6 Further Graphing of Quadratic Functions

2
Section 8.3

• Steps in solving quadratic equations
• If the equation is in the form (axb)2 c, use
  the square root property to solve.
• If not solved in step 1, write the equation in
  standard form.
• Try to solve by factoring.
• If you havent solved it yet, use the quadratic
  formula.

3

• Note that the directions on the previous slide
  did NOT include completing the square.
• Completing the square often involves more
  complicated computations with fractions, which
  can be avoided by using the quadratic formula.

4

• Solve 12x 4x2 4.
• 0 4x2 12x 4
• 0 4(x2 3x 1)
• Let a 1, b -3, c 1

5

• Solve the following quadratic equation.

6

• The steps we detailed in solving quadratic
  equations will only work if the equation is in an
  obviously recognizable form.
• Sometimes, we may have to alter the form of an
  equation to get it into quadratic form.
• This might involve substitution into the
  equation, squaring both sides of the equation, or
  converting a rational equation.

7
Solve the following equation.

• We need to isolate the radical and square both
  sides of the equation.

8

• We now have the equation in a standard
  quadratic form, so we can use quadratic
  techniques to solve it. Try factoring the
  quadratic equation and solve the equation.

• Now we need to set each of the factors equal to
  0 and solve the equations.
• x 8 0 or x 2 0
• which gives us x 8 or x 2.

9

• Now we substitute the values for x into the
  original equation.

true
false
So the solution is x 8.
10
Solve the following equation.
We now have the equation in a standard quadratic
form, so we can use quadratic techniques to solve
it.
11
Let a 1, b -9, c -6
Now we substitute the values for x into the
original equation.
12
true
13
true
14
Since both values give true statements when
substituted into the original equation, both
values are solutions.
15

• Solve the following equation.
• 9x4 5x2 4 0
• Substitute w x2 into the equation.
• 9w2 5w 4 0
• Now factor the equation.
• (9w 4)(w 1) 0
• Substitute the original variable back into the
  equation.
• (9x2 4)(x2 1) 0 (3x 2)(3x 2)(x2
  1)

16

• Now we need to set each of the preceding
  factors equal to 0 and solve the equations.
• 3x 2 0 or 3x 2 0 or x2 1 0

• Now we substitute the values for x into the
  original equation.

17
true
true
9i4 5i2 4 0 9(1) 5(-1) 4 0
9(-i)4 5(-i)2 4 0 9(1) 5(-1) 4 0
true
true
18

• Solve the following equation.
• 3x2/3 11x1/3 4
• Substitute w x1/3 into the equation.
• 3w2 11w 4 0
• Now factor the equation.
• (3w 1)(w 4) 0
• Substitute the original variable back into the
  equation.
• (3x1/3 1)(x1/3 4) 0

19

• Now we need to set each of the preceding
  factors equal to 0 and solve the equations.
• 3x1/3 1 0 or x1/3 4 0
• Solving for the first equation, we get 3x1/3 1
  

20
Solving for the second equation, we get x1/3
-4 (x1/3)3 (-4)3
x -64

• Now we substitute the values for x into the
  original equation.

3(-64)2/3 11(-64)1/3 4
3(16) 11(-4) 4
true
true
21
Section 8.4

• We previously have solved linear inequalities.
• A quadratic inequality is an inequality that can
  be written so that one side is a quadratic
  expression and the other side is 0.

3x2 4 gt 0 -2x2 5x 7 ? 0
x2 4x 6 ? 0 2x2
3 lt 0
22

• A solution of a quadratic inequality is a value
  of the variable that makes the inequality a true
  statement.
• If we attempt to solve a quadratic inequality,
  such as
• 3x2 5x 2 lt 0
• we are looking for values of x that will make
  this a true statement.
• If we graph the quadratic equation y 3x2 5x
  2, the points of the parabola that lie below
  the x-axis would provide values of x where the
  y-value lt 0. Hence, those values of x would
  satisfy 3x2 5x 2 lt 0.
• Similarly, we could also use the graph to find
  the values of x that satisfy the inequality 3x2
  5x 2 gt 0 (the x-values of all points above the
  x-axis).

23

• The points on the graph above and below the
  x-axis are separated by points actually on the
  x-axis. These points would have values of x such
  that 3x2 5x 2 0.
• However, graphing a quadratic equation could be
  time consuming if you dont have a computer or
  graphing calculator.
• Fortunately, it is not necessary to graph a
  quadratic inequality to solve this type of
  problem.
• We can construct a number line representing the
  x-axis and find the region(s) on the number line
  where the inequality is true.

24

• Solve the quadratic inequality 3x2 5x 2 lt 0.
• First we need to know when 3x2 5x 2 0.
• Factor the quadratic equation.
• (3x 2)(x 1) 0
• So x -2/3 or x -1.
• We can then examine the three regions of the
  number line that are created by these two values.

25

• All the values of x within the same region will
  produce quadratic values of the same sign
  (positive or negative), since the sign of the
  values on the parabola cannot change without
  passing through a zero value.
• So we only need to test a single value within a
  region to find out the sign of ALL quadratic
  values within that region.

26
The values in region B satisfy the inequality,
but we cannot include the endpoints, since they
will produce values that make the inequality 0,
rather than simply lt 0. So the solution is the
set (-1, -2/3).
27

• The techniques we used in the previous example
  can be extended to polynomials of degree higher
  than 2, as well.
• Solving a Polynomial Inequality
• Write the inequality in standard form, then solve
  the related EQUATION ( 0).
• Separate the number line into regions with the
  solutions from the equation in step 1.
• For each region, choose a test point and
  determine whether its value satisfies the
  original inequality.
• The solution set includes all the regions whose
  test point value is a solution of the inequality.
  
• If the inequality symbol is ? or ?, we include
  the endpoints of the regions.
• If the inequality symbol is lt or gt, the endpoints
  are not included.

28

• Solve the polynomial inequality x(x 6)(x 2) ?
  0.
• First we need to know when x(x 6)(x 2) 0.
  
• Since the polynomial is already factored, we
  set each factor 0 and solve.
• x 0 or x 6 0 or x 2 0
• So x 0, x 6 or x -2.
• We can then examine the four regions of the
  number line that are created by these three
  values.

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30
The values in regions B and D satisfy the
inequality, but we must include the endpoints,
since they will produce values that make the
inequality 0, rather than simply gt 0. So the
solution is the set -2, 0 ? 6, ?)
31

• These techniques can be extended to
  inequalities that contain rational expressions,
  as well as simply polynomial expressions.
• In this case, however, not only do you have to
  find the values that solve the equation when is
  substituted for the inequality sign, but also
  values of x that must be excluded from the
  solution set because they make the denominator
  0.
• We have to remember to exclude values in the
  solution set that will give us a 0 denominator,
  regardless of the type of inequality sign used.

32

• First we need to know when the rational
  expression is undefined (denominator 0).
• x 10 0 ? x 10
• Then we need to solve the equality

33

• The equality will hold when x 10 0 ?
  x -10.
• We can then examine the three regions of the
  number line that are created by these two values.

34
The values in regions A and C satisfy the
inequality, but we cannot include the endpoint in
region C, even though the symbol is ?, since it
will give us a 0 denominator. So the solution is
the set (-?, -10 ? (10, ?).
35

• First we need to know when the rational
  expression is undefined (denominator 0).
• p 4 0 ? p -4
• Then we need to solve the equality

36

• Multiplying both sides of the equation by p
  4, the equality will hold when p 3p(p 4)
  ? p 3p2 12p ? 0 3p2 11p
• We can factor this quadratic to solve the
  equation.
• p(3p 11) 0 ? p 0 or p -11/3
• We can then examine the four regions of the
  number line that are created by these three
  values.

37
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38
The values in regions B and D satisfy the
inequality, but we cannot include the left
endpoint in region B, even though the symbol is
?, since it will give us a 0 denominator. So the
solution is the set (-4, -11/3 ? 0, ?).
39
Section 8.5
We first examined the graph of f(x) x2 back in
chapter 3. We looked at the graphs of general
quadratic functions of the form f(x) ax2 bx
c in chapter 5. We discovered that the graph of a
quadratic function is a parabola opening upward
or downward, depending on the coefficient of the
x2 term, a. The highest point or lowest point on
the parabola is the vertex. Axis of symmetry is
the line that runs through the vertex and through
the middle of the parabola.
40

• Graph f(x) x2
• Note that a 1 in standard form.
• Which way does it open?
• What is the vertex?
• What is the axis of symmetry?
• Graph g(x) x2 3 and h(x) x2 3
• What is the vertex of each function?
• What is the axis of symmetry of each function?

41
f(x) x2
g(x) x2 3
h(x) x2 3
42

• In general how does the graph of y x2 k
  compare with the graph of y x2?
• For each value of x in y x2 k, the parabola
  is shifted upward k units on the graph.
• In general how does the graph of y x2 k
  compare with the graph of y x2?
• For each value of x in y x2 k, the parabola
  is shifted downward k units on the graph.
• The vertex is at (0, k) and the axis of symmetry
  is the y-axis.

43

• Graph f(x) x2
• Graph g(x) (x 3)2 and h(x) (x 3)2
• What is the vertex of each function?
• What is the axis of symmetry of each function?

44
f(x) x2
g(x) (x 3)2
h(x) (x 3)2
45

• In general how does the graph of y (x h)2
  compare with the graph of y x2?
• The parabola is shifted to the right k units on
  the graph.
• In general how does the graph of y (x h)2
  compare with the graph of y x2?
• The parabola is shifted to the left k units on
  the graph.
• The vertex is at (h, 0) and the axis of symmetry
  is the vertical line x h.

46

• Note The graphs of functions of the form
  g(x) (x h)2 or h(x) (x h)2
  might seem to be shifted in the opposite
  direction of what you would expect.
• After all, we usually think of minus as moving
  to the left, and plus as moving to the right.
• However, if you rewrite the function h(x) (x
  h)2 as h(x) (x (-h))2 and focus on whether
  you are subtracting a positive number (shift to
  the right) or subtracting a negative number
  (shift to the left), then the sign of the number
  being subtracted from x matches the direction of
  the shift.

47

• Graph f(x) x2
• Graph g(x) (x 2)2 4
• What is the vertex?
• What is the axis of symmetry?

48
f(x) x2
g(x) (x 2)2 4
49

• In general how does the graph of
  y (x h)2 k compare with the graph of
  y x2?
• The graph is shifted horizontally h units.
• If h is positive, shift to the right
• If h is negative, shift to the left
• The graph is shifted vertically k units.
• If k is positive, shift upward
• If k is negative, shift downward
• The vertex is at (h, k) and the axis of symmetry
  is the vertical line x h.

50

• Graph f(x) x2
• Graph g(x) 3x2 and h(x) (1/3)x2
• How do the shapes of the graphs compare?

51
f(x) x2
g(x) 3x2
h(x) (1/3)x2
52

• In general how does the graph of y ax2
  compare with the graph of y x2
• when a 1
• They are identical.
• when a gt 1
• It is narrower than the standard graph.
• when a lt 1
• It is wider than the standard graph.
• The vertex is at (0, 0) and the axis of symmetry
  is the y-axis.

53
Without graphing, how does the graph of
g(x) -4(x 9)2 1 compare with the graph
of f(x) x2? First rewrite the function in the
form of g(x) -4(x (-9))2
1. This new format tells us that if we start with
the standard parabola, we shift 9 units to the
left and one unit down. Then we flip it so that
it opens down. The shape then becomes much
wider. The vertex will be at (-9, -1) and the
axis of symmetry is the vertical line x -9.
54
Section 8.6

• In the previous section, we discovered that if
  the equation of a quadratic is written in the
  right format, we know a lot of information about
  the graph
• If the quadratic is written in the form f(x)
  a(x h)2 k, then we can find the vertex, axis
  of symmetry, whether it opens up or down, and the
  width.

55

• Since we can find out a lot about a quadratic
  function before we ever graph it, it would be in
  our best interest to get the quadratic into the
  appropriate form so we can easily find that
  information.
• The techniques we use would be similar to
  completing the square.

56

• Graph f(x) -3x2 6x 4. Find the vertex,
  axis, and any intercepts.
• Before graphing, we rewrite the equation into the
  form that will communicate information about the
  graph.

57
The vertex is at (1,7).
The axis of symmetry is the line x 1
58

• Intercepts for f(x) -3(x 1)2 7 occur when
  x 0 or y 0.
• y-intercept occurs when we set x 0.
• y -3(0 1)2 7
• y -3(- 1)2 7
• y -3(1) 7
• y 4
• y-intercept is the point (0, 4).

59

• x-intercept(s) occur when we set y 0
• 0 -3(x 1)2 7
• 3(x 1)2 7
• (x 1)2 7/3

60
Graph f(x) -3(x 1)2 7.
The vertex is at (1,7).
The axis of symmetry is the line x 1.
y-intercept is the point (0,4).
61

• In many applications, you are not interested in
  the entire graph of the quadratic function, but
  merely the vertex (the highest or lowest point of
  the graph).
• In that case, it is not necessary to convert
  the equation of the quadratic into the previous
  form that gives you much more information than
  just the vertex.
• There is a formula (derived in the text, if you
  are interested) for finding the vertex of the
  parabola using standard form.

62

• The maximum or minimum value of the parabola
  occurs at this vertex.
• The maximum or minimum value will be the second
  coordinate of the vertex.

63

• The Utah Ski Club sells calendars to raise money.
  The profit P, in cents, from selling x calendars
  is given by the equation P(x) 360x x2. What
  is the maximum profit the club will earn?
• Since the maximum value will occur at the
  vertex, we find the coordinates according to the
  previous formula.
• a -1 and b 360, so

64
The maximum profit will occur when the club sells
180 calendars. We substitute that number into
the profit formula to find P(180) 360(180)
(180)2 64800 32400 32400 cents
(remember to read the problem carefully) 324