Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240

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Transcript Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240

Algebra 2 Interactive Chalkboard
Copyright © by The McGraw-Hill Companies, Inc.
Send all inquiries to:
GLENCOE DIVISION
Glencoe/McGraw-Hill
8787 Orion Place
Columbus, Ohio 43240
Lesson 8-1 Midpoint and Distance Formulas
Lesson 8-2 Parabolas
Lesson 8-3 Circles
Lesson 8-4 Ellipses
Lesson 8-5 Hyperbolas
Lesson 8-6 Conic Sections
Lesson 8-7 Solving Quadratic Systems
Example 1 Find a Midpoint
Example 2 Find the Distance Between Two Points
Example 3 Find the Farthest Point
Computers A graphing program draws a line
segment on a computer screen so that its ends
are at (5, 2) and (7, 8). What are the coordinates
of its midpoint?
or (6, 5)
Answer: The coordinates of the midpoint are (6, 5).
Find the midpoint of the segment with endpoints
at (3, 6) and (–1, –8).
Answer: (1, –1)
What is the distance between P(–1, 4) and Q(2, –3)?
Distance Formula
Let
and
Subtract.
or
Simplify.
Answer: The distance between the two points
is
units.
What is the distance between P(2, 3) and Q(–3, 1)?
Answer:
units
Multiple-Choice Test Item
Which point is farthest from (2, –3)?
A (0, 0)
B (3, 2)
C (–3, 0)
D (4, 1)
Read the Test Item
The word farthest refers to the greatest distance.
Solve the Test Item
Use the Distance Formula to find the distance from (2, –3)
to each point.
Distance to (0, 0)
Distance to (3, 2)
Distance to (–3, 0)
Distance to (4, 1)
The greatest distance is
units. So, the farthest
point from (2, –3) is (–3, 0).
Answer: C
Multiple-Choice Test Item
Which point is farthest from (2, 3)?
A (0, 0)
B (1, 2)
C (3, 4)
D (–2, 4)
Answer: D
Example 1 Analyze the Equation of a Parabola
Example 2 Graph Parabolas
Example 3 Graph an Equation Not in Standard Form
Example 4 Write and Graph an Equation for a Parabola
Write
in standard form. Identify
the vertex, axis of symmetry, and direction of
opening of the parabola.
Original equation
Factor –1 from
the x-terms.
Complete the square
on the right side.
The 1 you added when
you completed the square
is multiplied by –1.
Answer: The vertex of this parabola is located at (–1, 4)
and the equation of the axis of symmetry is
.
The parabola opens downward.
Write
in standard form. Identify
the vertex, axis of symmetry, and direction of
opening of the parabola.
Answer:
axis of symmetry:
Graph
.
For this equation,
and
The vertex is at the
origin. Since the equation of the axis of symmetry is
substitute some small positive integers for x and find the
corresponding y-values.
x
1
2
3
y
2
8
18
Since the graph is symmetric about the
y-axis, the points at (–1, 2), (–2, 8) and
(–3, 18) are also on the parabola. Use
all of these points to draw the graph.
Answer:
Graph
.
The equation is of the form
where
The graph of this equation is the graph
of
in part a translated 1 unit right and 5 units
down. The vertex is now at (1, –5).
Answer:
Graph each equation.
a.
b.
Answer:
Answer:
Graph
First write the equation in the form
There is a y2 term, so
isolate the y and y2 terms.
Complete the square.
Add and subtract 4,
since
Then use the following information to draw the graph.
vertex: (3, 2)
axis of symmetry:
focus:
or
directrix:
direction of opening: left, since
length of the latus rectum:
or 1 unit
Answer:
Graph
Answer:
Bridges The 52 meter-long Hulme Arch Bridge in
Manchester, England, is supported by cables
suspended from a parabolic steel arch. The highest
point of the arch is 25 meters above the bridge, and the
focus of the arch is about 18 meters above the bridge.
Let the bridge be the x-axis, and let the y-axis pass
through the vertex of the arch. Write an equation that
models the arch.
The vertex is at (0, 25), so
and
The focus is at (0, 12). Use the y-coordinate
of the focus to find a.
k = 25; The y-coordinate of the
focus is 18.
Subtract 25 from each side.
Multiply each side by 4a.
Divide each side by –28.
Answer: An equation of the parabola is
Graph the equation.
The length of the latus rectum is
or 28 units, so
the graph must pass through (–14, 18) and (14, 18).
According to the length of the bridge, the graph must
pass through the points (–26, 0) and (26, 0). Use these
points and the information from part a to draw the graph.
Answer:
Fountains An outdoor fountain has a jet through which
water flows. The water stream follows a parabolic path.
The highest point of the water stream is
feet above
the ground and the water hits the ground 10 feet from
the jet. The focus of the fountain is
feet above the
ground.
a. Write an equation that models the path of the water
fountain.
Answer:
b. Graph the equation.
Answer:
Example 1 Write an Equation Given the Center
and Radius
Example 2 Write an Equation Given a Diameter
Example 3 Write an Equation Given the Center and
a Tangent
Example 4 Graph an Equation in Standard Form
Example 5 Graph an Equation Not in Standard Form
Landscaping The plan for a park puts the center of a
circular pond, of radius 0.6 miles, 2.5 miles east and
3.8 miles south of the park headquarters. Write an
equation to represent the border of the pond, using
the headquarters as the origin.
Since the headquarters is at (0, 0), the center of the pond
is at (2.5, –3.8) with radius 0.6 mile.
Equation of a circle
Simplify.
Answer: The equation is
Landscaping The plan for a park puts the center of a
circular pond, of radius 0.5 mile, 3.5 miles west and
2.6 miles north of the park headquarters. Write an
equation to represent the border of the pond, using
the headquarters as the origin.
Answer:
Write an equation for a circle if the endpoints of the
diameter are at (2, 8) and (2, –2).
Explore
To write an equation for a circle, you must
know the center and the radius.
Plan
You can find the center of the circle by
finding the midpoint of the diameter.
Then you can find the radius of the circle
by finding the distance from the center to
one of the given points.
Solve
First, find the center of the circle.
Midpoint Formula
Add.
Simplify.
Now find the radius.
Distance Formula
Subtract.
Simplify.
The radius of the circle is 5 units, so
Substitute h, k, and r2 into the standard
form of the equation of a circle.
Answer: An equation of the circle is
Examine
Each of the given points satisfies the
equation, so the equation is reasonable.
Write an equation for a circle if the endpoints of the
diameter are at (3, 5) and (3, –7).
Answer:
Write an equation for a circle with center at (3, 5) that
is tangent to the y-axis.
Sketch the circle. Since it
is tangent to the y-axis,
the radius is 3.
Answer: An equation of this circle is
.
Write an equation for a circle with center at (2, 3) that
is tangent to the x-axis.
Answer:
Find the center and radius of the circle with equation
Then graph the circle.
Answer: The center is at (0, 0) and the radius is 4.
The table lists some values for x and y that satisfy
the equation.
x
y
0
4
1
2
3.9
3.5
3
4
2.6
0
Since the circle is centered at
the origin, it is symmetric about
the y-axis. Therefore, the points
at (–1, 3.9), (–2, 3.5), (–3, 2.6)
and (–4, 0) lie on the graph.
The circle is also symmetric about the x-axis, so the
points (–1, –3.9), (–2, –3.5), (–3, –2.6), (1, –3.9),
(2, –3.5), (3, –2.6), and (0, –4) lie on the graph.
Graph these points and draw the circle that
passes through them.
Answer:
Find the center and radius of the circle with equation
Then graph the circle.
Answer:
center (0, 0);
Find the center and radius of the circle with equation
Then graph the circle.
Complete the square.
(–3, 0)
Answer: The center is
at (–3, 0) and the
radius is 4.
Find the center and radius of the circle with equation
Then graph the circle.
Answer:
center (–4, 2);
(–4, 2)
Example 1 Write an Equation for a Graph
Example 2 Write an Equation Given the Lengths of
the Axes
Example 3 Graph an Equation in Standard Form
Example 4 Graph an Equation Not in Standard Form
Write an equation for the ellipse shown.
In order to write an
equation for the ellipse,
we need to find the
values of a and b for
the ellipse. We know
that the length of the
major axis of any ellipse
is 2a units. In this
ellipse, the length of the
major axis is the
distance between (0, 5)
and (0, –5). This
distance is 10 units.
Divide each side by 2.
The foci are located at (0, 4) and (0, –4), so c = 4. We
can use the relationship between a, b, and c to
determine the value of b.
Equation relating a, b, and c
and
Solve for b2.
Since the major axis is vertical, substitute 25 for a2 and
9 for b2 in the form
Answer: An equation of the ellipse is
Write an equation for the ellipse shown.
Answer:
Sound A listener is standing in an elliptical room 150
feet wide and 320 feet long. When a speaker stands at
one focus and whispers, the best place for the listener
to stand is at the other focus.
Write an equation to model this ellipse, assuming the
major axis is horizontal and the center is at the origin.
The length of the major axis is 320 feet.
Divide each side by 2.
The length of the minor axis is 150 feet.
Divide each side by 2.
Substitute
and
into the form
Answer: An equation for the ellipse is
How far apart should the speaker and the listener
be in this room?
The two people should stand at the two foci of the ellipse.
The distance between the foci is 2c units.
Equation relating a, b, and c
Take the square root
of each side.
Multiply each side by 2.
Substitute
and
Use a calculator.
Answer: The two people should be about 282.7 feet apart.
Sound A listener is standing in an elliptical room 60
feet wide and 120 feet long. When a speaker stands at
one focus and whispers, the best place for the listener
to stand is at the other focus.
a. Write an equation to model this ellipse, assuming the
major axis is horizontal and the center is at the origin.
Answer:
b. How far apart should the speaker and the listener be
in this room?
Answer: 103.9 feet apart
Find the coordinates of the center and foci and the
lengths of the major and minor axes of the ellipse with
equation
Then graph the equation.
The center of this ellipse is at (0, 0).
Since
and since
The length of the major axis is 2(6) or 12 units, and the
length of the minor axis is 2(3) or 6. Since the x2 term has
the greatest denominator, the major axis is horizontal.
Equation relating a, b, and c
Take the square root of each side.
The foci are at
and
You can use a calculator
to find some approximate
nonnegative values for x
and y that satisfy the equation.
Since the ellipse is centered at
the origin, it is symmetric about
the y-axis. So, the points at
(1, 2.96) and (–1, 2.96) lie on
the graph.
The ellipse is also symmetric about
the x-axis, so the points at (1, –2.96)
and (–1, –2.96) also lie on the graph.
x
y
0
3
1
2
2.96
2.83
3
4
5
6
2.60
2.24
1.66
0
Graph the intercepts (–6, 0) (6, 0) (0, 3) and (0, –3)
and draw the ellipse that passes through them and
the other points.
Answer:
center: (0, 0);
foci:
major axis: 12;
minor axis: 6
Find the coordinates of the center and foci and the
lengths of the major and minor axes of the ellipse with
equation
Answer:
center: (0, 0);
foci:
major axis: 10;
minor axis: 4
Then graph the equation.
Find the coordinates of the center and foci and the
lengths of the major and minor axes of the ellipse
with equation
Then graph the ellipse.
Complete the square to write in standard form.
Original equation
Complete the squares.
Write the trinomials
as perfect squares.
Divide each side
by 36.
Answer: The center is (3, 2) and the foci are located at
and
The length of the
major axis is 12 units and the length of the minor
axis is 6.
Find the coordinates of the center and foci and the
lengths of the major and minor axes of the ellipse
with equation
Then graph the ellipse.
Answer:
center: (–2, 3);
foci:
major axis: 10;
minor axis: 4
Example 1 Write an Equation for a Graph
Example 2 Write an Equation Given the Foci and
Transverse Axis
Example 3 Graph an Equation in Standard Form
Example 4 Graph an Equation Not in Standard Form
Write an equation for the hyperbola.
The center is the midpoint of
the segment connecting the
vertices, or (0, 0).
The value of a is the
distance from the center
to a vertex or 2 units.
The value of c is the
distance from the center
to a focus, or 4 units.
Equation relating a, b,
and c for a hyperbola
Evaluate the squares.
Solve for b2.
Since the transverse axis is vertical, the equation is of
the form
Substitute the values for a2 and b2.
Answer: An equation of the hyperbola is
Write an equation for the hyperbola.
Answer:
Navigation A ship notes that the difference of its
distance from two LORAN stations that are located at
(–70, 0) and (70, 0) is 70 nautical miles. Write an
equation for the hyperbola on which the ship lies.
First draw a figure.
By studying either of the x-intercepts, you can see that the
difference of the distances from any point on the hyperbola
to the stations at the foci is the same as the length of the
transverse axis, or 2a. Therefore,
or
According to the coordinates of the foci,
Use the values for a and c to find b for this hyperbola.
Equation relating a, b,
and c for a hyperbola
Evaluate the squares.
Solve for b2.
Since the transverse axis is horizontal, the equation is of
the form
Substitute the values for a2 and b2.
Answer: An equation of the hyperbola is
Navigation A ship notes
that the difference of its
distance from two LORAN
stations that are located at
(–60, 0) and (60, 0) is 60
nautical miles. Write an
equation for the hyperbola
on which the ship lies.
Answer:
Find the coordinates of the vertices and foci and the
equations of the asymptotes for the hyperbola with
equation
Then graph the hyperbola.
Answer: The center of the hyperbola is at the origin.
According to the equation,
and
so
and
The coordinates of the vertices are (1, 0)
and (–1, 0).
Equation relating a, b,
and c for a hyperbola
Simplify.
Take the square
root of each side.
Answer: The foci are at
and
The equations of the asymptotes are
or
or
You can use a calculator to
find some approximate
nonnegative values for x and y
that satisfy the equation.
x
y
1
2
3
4
5
0
1.7
2.8
3.9
4.9
Since the hyperbola is centered at the origin, it is
symmetric about the y-axis. Therefore, the points at
(–5, 4.9), (–4, 3.9), (–3, 2.8), (–2, 1.7), and (–1, 0)
lie on the graph.
The hyperbola is also symmetric about the x-axis, so the
points at (–5, –4.9), (–4, –3.9), (–3, –2.8), (–2, –1.7),
(2, –1.7), (3, –2.8), (4, –3.9), and (5, –4.9) also lie on
the graph.
Draw a 2-unit by 2-unit square. The asymptotes contain
the diagonals of the square. Graph the vertices, which,
in this case, are the x-intercepts. Use the asymptotes as
a guide to draw the hyperbola that passes through the
vertices and the other points. The graph does not
intersect the asymptotes.
Answer:
Find the coordinates of the vertices and foci and the
equations of the asymptotes for the hyperbola with
equation
Then graph the hyperbola.
Answer:
vertices: (2, 0), (–2, 0);
foci:
asymptotes:
Find the coordinates of the vertices and foci and the
equations of the asymptotes for the hyperbola with
equation
Then graph the hyperbola.
Complete the square for each variable to write in
standard form.
Original equation
Complete the square.
Write the trinomials
as perfect squares.
Answer: The vertices are (–4, 5) and (–2, 5) and the
foci are
the asymptotes are
and
and
The equations of
or
Find the coordinates of the vertices and foci and the
equations of the asymptotes for the hyperbola with
equation
Then graph the hyperbola.
Answer:
vertices: (4, 1), (4, –5);
foci: (4, 3), (4, –7);
asymptotes:
Example 1 Rewrite an Equation of a Conic Section
Example 2 Analyze an Equation of a Conic Section
Write the equation
in standard form.
State whether the graph of the equation is a parabola,
circle, ellipse, or hyperbola. Then graph the equation.
Write the equation is standard form.
Original equation
Isolate terms.
Divide each side by 18.
Answer: The graph is an ellipse with center at (0, 0).
Write the equation
in standard
form. State whether the graph of the equation is a
parabola, circle, ellipse, or hyperbola. Then graph
the equation.
Answer:
circle
Without writing the equation in standard form, state
whether the graph of
is a parabola,
circle, ellipse, or hyperbola.
Answer:
and
Since A and C have opposite signs,
the graph is a hyperbola.
Without writing the equation in standard form, state
whether the graph of
is a parabola, circle, ellipse, or hyperbola.
Answer:
and
Since
the graph is a circle.
Without writing the equation in standard form, state
whether the graph of
is a parabola,
circle, ellipse, or hyperbola.
Answer:
and
Since
this graph is a parabola.
Without writing the equation in standard form, state
whether the graph of the equation is a parabola, circle,
ellipse, or hyperbola.
a.
Answer: hyperbola
b.
Answer: ellipse
c.
Answer: parabola
Example 1 Linear-Quadratic System
Example 2 Quadratic-Quadratic System
Example 3 System of Quadratic Inequalities
Solve the system of equations.
You can use a graphing calculator to help visualize the
relationships of the graphs of the equations and predict
the number of solutions.
Solve each equation for y to obtain
and
Enter the functions on the Y= screen. The graph
indicates that the hyperbola and the line intersect
in one point. So, the system has one solution.
Use substitution to solve the system.
First, rewrite
First equation in the system
Substitute 2 – 2y for x.
Simplify.
Subtract 16 from each side.
Divide each side by –32.
Now solve for x.
Equation for x in terms of y
Substitute the y value.
Simplify.
Answer: The solution is
Solve the system of equations.
Answer: (2, 0) and
Solve the system of equations.
A graphing calculator
indicates that the
circle and ellipse
intersect in four
points. So, this
system has four
solutions.
Use the elimination method to solve the system.
Rewrite the first
original equation.
Second original equation
Add.
Divide each side by 3.
Take the square
root of each side.
Substitute
and
in either of the original
equations and solve for y.
Original equation
Substitute for x.
Subtract
each side.
from
Take the square
root of each side.
Answer: The solutions are
and
Solve the system of equations.
Answer: (3, 1), (3, –1), (–3, 1), and (–3, –1)
Solve the system of inequalities by graphing.
The graph of
is the parabola
and the
region inside and above it. The region is shaded blue.
The graph of
is the interior of the circle
This region is shaded yellow.
Answer:
The intersection of these regions, shaded green,
represents the solution of the system of inequalities.
Solve the system of inequalities by graphing.
Answer:
Explore online information about the
information introduced in this chapter.
Click on the Connect button to launch your browser
and go to the Algebra 2 Web site. At this site, you
will find extra examples for each lesson in the
Student Edition of your textbook. When you finish
exploring, exit the browser program to return to this
presentation. If you experience difficulty connecting
to the Web site, manually launch your Web browser
and go to www.algebra2.com/extra_examples.
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