Intermediate Algebra: A Graphing Approach

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Intermediate Algebra A Graphing Approach

• 10.1 The Parabola Circle
• 10.2 The Ellipse Hyperbola
• 10.3 Solving Nonlinear Systems of Equations
• 10.4 Nonlinear Inequalities Systems of
  Inequalities

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Section 10.1 The Parabola and the Circle

• Conic sections derive their name because each
  conic section is the intersection of a right
  circular cone and a plane.

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• Just as y a(x h)2 k is the equation of a
  parabola that opens upward or downward, x a(y
  k)2 h is the equation of a parabola that opens
  to the right or to the left.

y a(x h)2 k
x a(y k)2 h
a gt 0
(h, k)
(h, k)
(h, k)
y k
a lt 0
y k
(h, k)
a lt 0
a gt 0
x h
x h
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Graph the parabola x (y 4)2 1.

• a gt 0, so the parabola opens to the right.
• The vertex of the parabola is (1, 4).
• The axis of symmetry is y 4.

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The table shows ordered pairs of the solutions of
x (y 4)2 1.
1
4
y 4
2
3
2
5
17
0
17
8
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Graph the parabola y x2 12x 25.

• Complete the square on x to write the equation in
  standard form.

y 25 x2 12x
Subtract 25 from both sides.

• The coefficient of x is 12. The square of half
  of 12 is 62 36.

y 25 36 x2 12x 36
Add 36 to both sides.
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y 11 (x 6)2
Simplify the left side and factor the right side.
y (x 6)2 11
Subtract 11 from both sides.

• a gt 0, so the parabola opens upward.
• The vertex of the parabola is ( 6, 11).
• The axis of symmetry is x 6.

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y x2 12x 25
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Find the distance between ( 6, 6) and ( 5,
2).
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• The midpoint of a line segment is the point
  located exactly halfway between the two endpoints
  of the line segment.

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Find the midpoint of the line segment that joins
points P(0, 8) and Q(4, 6).
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• A circle is the set of all points in a plane that
  are the same distance from a fixed point called
  the center. The distance is called the radius.

Circle The graph of (x h)2 (y k)2 r2 is a
circle with center (h, k) and radius r.
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Graph (x 3)2 y2 9.

• The equation can be written as (x 3)2 (y
  0)2 32.
• h 3, k 0, and r 3.

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Find the equation of the circle with center ( 7,
6) and radius 2.

• h 7, k 6, and r 2.
• (x h)2 (y k)2 r2.

The equation can be written as x ( 7)2 (y
6)2 22.
(x 7)2 (y 6)2 4.
Simplify.
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Section 10.2 The Ellipse and the Hyperbola

• An ellipse can be thought of as the set of points
  in a plane such that the sum of the distances of
  those points from two fixed points is constant.
• Each of the two fixed points is called a focus.
  (The plural of focus is foci.)
• The point midway between the foci is called the
  center.

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• The equation is of the form
• a 2 and b 3.

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Hyperbola with Center (0, 0)
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• The asymptotes of the hyperbola are dashed lines
  used to sketch the graph of the hyperbola.
• To sketch the asymptotes, draw a rectangle with
  vertices (a, b), ( a, b), (a, b), and ( a,
  b).

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• It has center (0, 0) and x-intercepts (3, 0) and
  ( 3, 0).

• The asymptotes of the hyperbola are the extended
  diagonals of the rectangle with corners (3, 6),
  ( 3, 6), (3, 6), and ( 3, 6 ).

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Section 10.3 Solving Nonlinear Systems of
Equations

• A nonlinear system of equations is a system of
  equations where at least one of the equations is
  not linear.
• The substitution method or the elimination method
  may be used to solve the system.

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Solve the system.

• The substitution method will work best to solve
  this system.

y 2 x
Second equation
x2 y 4
First equation
(y 2)2 y 4
Replace x with y 2.
y2 4y 4 y 4
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y2 4y 4 y 4
y2 5y 0
y(y 5) 0
y 0 or y 5

• Let y 0 and then let y 5 to find the
  corresponding x values.

Let y 0
Let y 5
y 2 x
y 2 x
0 2 x
5 2 x
x 2
x 3
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• The solutions are (2, 0) and ( 3, 5).
• Check both solutions in both equations.

y 2 x
x2 y 4
y 2 x
x2 y 4
( 3)2 ( 5) 4
5 2 3
22 0 4
0 2 2
9 5 4
3 3
4 4
2 2
y 2 x

• A graph of the system also verifies the solution.

x2 y 4
(2, 0)
( 3, 5).
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Solve the system.

• The elimination method will work best to solve
  this system.

( 1) x2 ( 1) 2y2 ( 1) 4
Multiply the first equation by 1.
x2 2y2 4
x2 y2 4
3y2 0
Add the two equations.
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3y2 0
y 0

• Let y 0 to find the corresponding x values.

x2 2y2 4
x2 2(0)2 4
x2 4
x 2 or x 2

• The solutions are ( 2, 0) and (2, 0).
• Check both solutions in both equations.

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Section 10.4 Nonlinear Inequalities and Systems
of Inequalities

• Nonlinear inequalities in two variables are
  graphed in a similar way to linear inequalities
  in two variables.

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• Select a test point to determine which region
  contains the solutions.
• Select the test point (0, 0) if it is not on the
  boundary line.

• This is a true statement, so the test point is
  part of the solution.

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Region A

• The hyperbola divides the plane into three
  regions. Select a test point in each region to
  determine the solutions.

Region B
Region C
Region A
Region B
Region C
False
True
False
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• The graph of the solution set includes the shaded
  region B only. It also includes the boundary
  lines.

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Graph the system.

• Graph each inequality on the same set of axes.

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• Shade the solution set for each inequality.

• The solution to the system is the dark green area
  where both regions are shaded.