Lesson 6.5, For use with pages 446-451 1.f(x) = 2x – 5. ANSWER domain and range: all real numbers...
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Lesson 6.5, For use with pages 446-451 1. f(x) = 2x – 5. ANSWER ANSWER domain and range: all real numbers 2. g(x) = – + 6 State the domain and range of the function. domain: all real numbers: range: y ≤ 6
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Transcript of Lesson 6.5, For use with pages 446-451 1.f(x) = 2x – 5. ANSWER domain and range: all real numbers...
Lesson 6.5, For use with pages 446-451
1. f(x) = 2x – 5.
ANSWER
ANSWER
domain and range: all real numbers
2. g(x) = – + 6
State the domain and range of the function.
domain: all real numbers: range: y ≤ 6
ANSWER
State the domain and range of the function.
2 3. y = 8.1 –
domain: all real numbers: range: y ≤ –8
Lesson 6.5, For use with pages 446-451
Alg II 6.5 PPGraphing Square Roots
(Radicals)
EXAMPLE 1 Graph a square root function
Graph y = and state the domain and range.
Compare the graph with the graph of y =
12
SOLUTION
Make a table of values and sketch the graph.
EXAMPLE 1 Graph a square root function
The radicand of a square root must be nonnegative.
So, the domain is x ≥ 0. The range is y ≥ 0.
The graph of y = is a vertical shrink of the graph
of y = by a factor of .
12
x12
EXAMPLE 2 Graph a cube root function
Graph y = –3 , and state the domain and range. Compare the graph with the graph of y = .
SOLUTION
Make a table of values and sketch the graph.
EXAMPLE 2 Graph a cube root function
The domain and range are all real numbers.
The graph of y = – 3 is a vertical stretch of the
graph of y = by a factor of 3 followed by a
reflection in the x-axis.
x3
x3
EXAMPLE 3 Solve a multi-step problem
• Use a graphing calculator to
graph the model.
• How long is a pendulum
with a period of 3 seconds?
Pendulums
The period of a pendulum is the time the pendulum takes to complete one back-and-forth swing. The period T (in seconds) can be modeled by T = 1.11 where l is the pendulum’s length (in feet).
EXAMPLE 3 Solve a multi-step problem
SOLUTION
STEP 1
Graph the model. Enter the equation y = 1.11 . The graph is shown below.
Use the trace feature to find the value of x when y = 3.The graph shows x 7.3.
STEP 2
EXAMPLE 3 Solve a multi-step problem
ANSWER
A pendulum with a period of 3 seconds is about 7.3 feet long.
GUIDED PRACTICE for Examples 1, 2 and 3
1. y = – 3
SOLUTION
ANSWER
Domain: Range:x > 0 y < 0
Make a table of values and sketch the graph.
GUIDED PRACTICE for Examples 1, 2 and 3
SOLUTION
ANSWER
Domain: Range:x > 0 y < 0
2. y =14
x
Make a table of values and sketch the graph.
x 1 2 3 4
y 0.25 0.35 0.43 0.5
GUIDED PRACTICE for Examples 1, 2 and 3
3. y = x312
–
ANSWER
Domain: Range:All real numbers All real numbers
x -3 -2 -1 0 1 2 3
y 0.72 0.63 0.5 0 -0.5 -0.6 -0.7
SOLUTION
Make a table of values and sketch the graph.
GUIDED PRACTICE for Examples 1, 2 and 3
4. g(x) = 4
ANSWER
Domain: Range:All real numbers All real numbers
x -3 -2 -1 0 1 2 3
y -5.77 -5.04 -4 0 4 5.04 5.77
SOLUTION
Make a table of values and sketch the graph.
3√𝑥
GUIDED PRACTICE for Examples 1, 2 and 3
5. What if ? Use the model in example 3 to find the length of a pendulum with a period of 1 second.
ANSWER
About 0.8 ft
EXAMPLE 4 Graph a translated square root function
Graph y = -2 +2. Then state the domain and range.
SOLUTION
STEP 1
Sketch the graph of y = –2 (shown in blue). Notice that it begins at the origin and passes through the point (1, –2).
EXAMPLE 4 Graph a translated square root function
STEP 2
Translate the graph. For y = - 2 +2 , h = 3 and k = 2. So, shift the graph of y = –2 right 3 units and up 2 units. The resulting graph starts at (3, 2) and passes through (4, 0).
From the graph, you can see that the domain of the function is x ≥ 3 and the range of the function is y ≤ 2.
EXAMPLE 5 Graph a translated cube root function
Graph y = 3 3
x + 4 – 1. Then state the domain and range.
SOLUTION
STEP 1
Sketch the graph of y = 3 3 x
(shown in blue). Notice that it begins at the origin and passes through the point (–1, –3) and (1, 3).
EXAMPLE 5 Graph a translated cube root function
STEP 2
Translate the graph. Note the for y = –3 3 x + 4 – 1, h = –4 and k = –1.
So, shift the graph of y = 3 3 x left 4 units and
up 1 unit. The resulting graph starts at (–5, –4),(–4, –1) and passes through (–3, 2).
From the graph, you can see that the domain and range of the function are both all real numbers.
GUIDED PRACTICE for Examples 4 and 5
6. y = –4 x + 2
ANSWER
Domain : x > 0 ,range : y < 0.
Graph the function. Then state the domain and range.
GUIDED PRACTICE for Examples 4 and 5
ANSWER
Domain : x > –1 ,range : y > 0.
7. y = x + 12
Graph the function. Then state the domain and range.
GUIDED PRACTICE for Examples 4 and 5
ANSWER
Domain : x > 3 ,range : y > –1.
8. f(x) = x – 3 – 12
1
Graph the function. Then state the domain and range.
GUIDED PRACTICE for Examples 4 and 5
ANSWER
Domain :all real numbers, range: all real numbers.
y = 9. x – 42 3
Graph the function. Then state the domain and range.
GUIDED PRACTICE for Examples 4 and 5
ANSWER
Domain :all real numbers, range: all real numbers.
y = 10. x – 53
Graph the function. Then state the domain and range.
GUIDED PRACTICE for Examples 4 and 5
ANSWER
Domain :all real numbers, range: all real numbers.
11. g(x) = x + 2 – 3
3
Graph the function. Then state the domain and range.
