**Unformatted text preview: **12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST WEEK 5 RATIONAL FUNCTIONS AND CONICS 1 - SECTIONS 3.1 - 3.2
3.1 Graphing Rational Functions - PRACTICE TEST - Grade Report
Score: 100% (22 of 22 pts) Submitted: Dec 2 at 1:29pm 1/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 1 Grade: 1.0 / 1.0 Which of the graphs is given by the following function?
3
x −27
f (x) =
2
x −x−6 Your response Solution To graph the function, write it in its simplest form. Simplify the function by factoring, then remove the common factors.
3
x −27
f (x)=
2
x −x−6
2
(x−3)(x
+ 3x + 9)
= Factor
(x−3)(x + 2)
2
x
+ 3x + 9 = Simplify
x + 2 The factor (x−3) was removed from the denominator. To find the x -coordinate of the hole set the factor equal to 0 and solve.
x−3 = 0; x = 3. Substitute 3 into the simplest form of the function to find the y -coordinate of the hole.
2
(3)
+ 3(3) + 9
g(3) = 27
= 3 + 2 5 To find the vertical asymptote, set the denominator equal to 0 and solve for x.
x + 2 = 0; x = −2. vertical asymptote: x = −2
There is no horizontal asymptotesince the degree of the numerator is greater than the degree of the denominator.
The degree of the numerator is one more than the degree of the denominator, so the graph has an oblique asymptote. Find the oblique asymptote
by using long division to divide the numerator by the denominator. 2/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
x+1 + 7
x + 2 x + 2 ¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
2
)x
+ 3x + 9
2
−(x
+ 2x)
¯
¯¯¯¯¯¯¯¯¯¯¯¯¯
x + 9
−(x+2)
¯
¯¯¯¯
¯¯¯¯
7 oblique asymptote: y = x+1
Substitute x -values from either side of the vertical asymptote into the simplified function and simplify to find the corresponding y -coordinate
of points on the graph. Plot these to graph the function. 3/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 2 Grade: 1.0 / 1.0 Find the coordinates of the hole, if it exists, in the graph of the following function.
3
2
x −4x
+ 4x−16
f (x) =
2
x −5x+4 Enter your answer as a coordinate pair including the comma. If a coordinate is not an integer, enter it as a fraction in simplest form. If the graph
does not have a hole, enter ”None”.
The coordinates of the hole are (4,20/3) (100%).
Solution To determine if the graph has a hole, write in simplest form.
3
2
x −4x
+ 4x − 16
f (x)=
2
x −5x+4
2
(x−4)(x
+ 4)
= Factor
(x − 4)(x−1)
2
(x
+ 4) = Remove the common factor
(x−1) The factor (x − 4) was removed from the denominator. Set this factor equal to 0 and solve for x to find the coordinate of the hole.
x−4 =0
x =4 Therefore, the graph contains a hole where x = 4.
Substitute 4 into the simplest form of the function to find the y -coordinate of the hole.
2
(4)
+ 4
f (x) = 20
= (4)−1 .
3 Thus, the graph contains a hole at (4, 20 ∕ 3). 4/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 3 Grade: 1.0 / 1.0 Find the asymptotes of the following function.
3
x −8
f (x) =
2
x +x−6 Complete the equation of each type of asymptote. If there are multiple asymptotes of a type, separate the answers with semicolons (;). If there is
no asymptote of that type, enter "none".
The vertical asymptote is at x = -3 (33%).
The horizontal asymptote is at y = none (33%).
The equation of the oblique asymptote is y = x-1 (33%). Solution To graph the function, write it in its simplest form. Simplify the function by factoring, then remove the common factors.
3
x −8
f (x)=
2
x +x−6
2
(x−2)(x
+ 2x + 4)
= Factor
(x−2)(x + 3)
2
x
+ 2x + 4 = Simplify
x + 3 The factor (x−2) was removed from the denominator. To find the x -coordinate of the hole set the factor equal to 0 and solve.
x−2 = 0; x = 2. Substitute 2 into the simplest form of the function to find the y -coordinate of the hole.
2
(2)
+ 2(2) + 4
g(2) = 12
= 2 + 3 5 To find the vertical asymptote, set the denominator equal to 0 and solve for x.
x + 3 = 0; x = −3. vertical asymptote: x = −3
There is no horizontal asymptotesince the degree of the numerator is greater than the degree of the denominator, so the answer is "none".
The degree of the numerator is one more than the degree of the denominator, so the graph has an oblique asymptote. Find the oblique asymptote
by using long division to divide the numerator by the denominator.
7
x−1 +
x + 3
x + 3 ¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
2
)x
+ 2x + 4
2
−(x
+ 3x)
¯
¯¯¯¯¯¯¯¯¯¯¯¯¯
−x + 4
−(−x−3)
¯¯¯¯¯¯¯¯¯¯¯
7 oblique asymptote: y = x−1 5/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST Substitute x -values from either side of the vertical asymptote into the simplified function and simplify to find the corresponding y -coordinate
of points on the graph. Plot these to graph the function. 6/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 4 Grade: 1.0 / 1.0 Which of the graphs is given by the following function?
2
2x −5x−3
f (x) =
x−3 Your response Solution To graph the function, write it in its simplest form. Simplify the function by factoring, then remove the common factors.
2
2x −5x−3
f (x) =
x − 3
(2x + 1)(x−3)
= Factor.
x−3 = 2x + 1 Remove the common factor. The factor (x − 3) was removed from the denominator. To find the x -coordinate of the hole, set the factor equal to 0 and solve. x−3 x 3 = 0; = Substitute 3 into the expression to find the y -coordinate of the hole.
f (3) = 2(3) + 1 = 7 Thus, the graph contains a hole at (3, 7). 7/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 5 Grade: 1.0 / 1.0 Find the coordinates of the hole, if it exists, in the graph of the following function.
2
3x −9x−12
f (x) =
x−4 Enter your answer as a coordinate pair, including the comma. If a coordinate pair is not an integer, enter it as a fraction in its simplest form. If
the graph does not have a hole, enter ”None”.
The coordinates of the hole are (4,15) (100%)
Solution To determine if the graph has a hole, write the function in simplest form.
2
3x −9x−12
f (x) =
x−4
(3x + 3)(x−4)
= Factor.
x−4 = 3x + 3 Remove the common factor. The factor (x − 4) was removed from the denominator. Set this factor equal to 0 and solve for x to find the coordinate of the hole.
x−4 = 0
x = 4 Therefore, the graph contains a hole where x = 4.
Substitute 4 into the expression to find the y -coordinate of the hole.
f (4) = 3(4) + 3 = 15 Thus, the graph contains a hole at (4, 15). 8/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 6 Grade: 1.0 / 1.0 Choose the function that best corresponds with the graph below. Your response y = 2
x −6x+5
x−3 Solution The graph has a vertical asymptote at x = 3, so the denominator must be x−3.
The graph has an oblique asymptote, so the numerator’s degree must be equal to the denominator's degree + 1.
Therefore, the best choice for the function that describes the given graph is
2
x −6x+5
y =
x−3 Check by finding the y -intercept and the equation of the oblique asymptote.
On the graph, the y -intercept is approximately(0, −1.67) . To find the y -intercept, Substitute 0 for x in the equation to find the y -intercept
from the equation.
2
(0) −6(0)+5
y = = −1.67,
(0)−3 which is the same as for the graph. Thus the y -intercept is at (0, −1.67).
On the graph, the equation of the oblique asymptote, shown in blue, is y = x−3 (since m = 1 and b = −3 ).
Alternatively, use long division to find the equation of the oblique asymptote. Dividing the numerator by the denominator gives a quotient x−3
with remainder −4 .
Thererfore, the equation of the oblique asymptote is y = x−3 , which is the same for the graph. 9/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 7 Grade: 1.0 / 1.0 Find the asymptotes for the following function.
2
3x +10x+9
f (x) =
x+2 Complete the equation of each type of asymptote. If there are multiple asymptotes of a type, separate the answers with semicolons (;). If there is
no asymptote of that type, enter "none".
The vertical asymptote(s) are at x = -2 (33%)
The horizontal asymptote is at y = none (33%)
The equatin of the oblique asymptote is y = 3x+4 (33%) Solution 2
3x +10x+9
f (x) = .
x+2 The function is in its simplest form.
To find the vertical asymptote, set the denominator equal to 0 and solve for x .
x+2 = 0; x = −2. vertical asymptote: x = −2
There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator.
The degree of the numerator is one more than the degree of the denominator, so the graph has an oblique asymptote. Find the oblique asymptote
by using long division to divide the numerator by the denominator.
1
3x + 4 +
x+2
x+2 ¯¯¯¯
¯¯¯¯¯¯¯¯¯¯¯¯¯
2
)3x +10x+9
2
−(3x +6x)
¯
¯¯¯¯¯¯¯¯¯¯¯¯¯
4x+9
−(4x+8)
¯
¯¯¯¯¯¯¯¯¯¯¯¯
1 oblique asymptote: y = 3x + 4 10/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 8 Grade: 1.0 / 1.0 Which of the graphs is given by the following function?
2
9x −54x−63
f (x) =
2
x
+ 2x−15 Choose the letter that corresponds to the correct graph.
A (100%) A. B. C. D. Solution Factor the numerator and denominator.
f (x) = 2
9x −54x−63
2
x
+ 2x−15
9(x − 7)(x + 1) =
(x − 3)(x + 5) Find the vertical asymptotes.
Set (x − 3)(x + 5) = 0, then x = 3 and x = −5. Find the horizontal asymptote.
11/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST The degree of the numerator (2) is equal to the degree of the denominator (2), so the function has a horizontal asymptote at
leading coefficient of the numerator
y = 9
= leading coefficient of the denominator = 9.
1 Find the zeros.
Set (x − 7)(x + 1) = 0, then x = 7 and x = −1.
Therefore, the function has x-intercepts at x = 7 and x = −1 . Find the y-intercept.
9(0 − 7)(0 + 1)
f (0) = 21
= (0 − 3)(0 + 5) 5 Therefore, the function has a y-intercept of (0, 21
5 ). A. So, the correct graph is 12/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 9 Grade: 1.0 / 1.0 Find the horizontal asymptote(s), if any, of
2x −3
f (x) = .
2
5x Enter the answer in slope-intercept form, x = c or y = b as applicable. If any number in the answer is not an integer, use the fraction template to
enter it as a fraction in simplest form. Write addition of a negative number as subtraction. If there are no horizontal asymptotes, enter ”None”.
The equation of the horizontal asymptote is y=0 (100%)
Solution The degree of the numerator (1) is less than the degree of the denominator (2), so the function has a horizontal asymptote at y Question: 10 = 0. Grade: 1.0 / 1.0 Choose the phrase that correctly completes the statement.
If r(x) is a rational function in simplest form where the degree of the numerator is 3 and the degree of the denominator is 1, then
r(x) has no horizontal asymptote (100%)
.
Solution There are three cases to consider; the degree of the numerator being greater, equal to, or less than the degree of the denominator.
1. If the degree of the numerator is greater than the degree of the denominator, then there is no horizontal asymptote.
2. If the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at y =
3. If the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at 0. leading coefficient of numerator
y = .
leading coefficient of denominator Here, the degree of the numerator, 3, is greater than the degree of the denominator, 1, so r(x) has no horizontal asymptote. 13/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 11 Grade: 1.0 / 1.0 Which of the graphs is given by the following function?
2x 2 −10x−12 f (x) =
x−2 Your response Solution The function can be factored as
2(x−6)(x+1)
f (x) = x−2 Find the vertical asymptotes.
Set x − 2 = 0,
then x = 2. Find the zeros.
Set (x − 6)(x + 1) = 0,
then x = 6 and x = −1. Therefore, the function has x -intercepts at x = 6 and x = −1. Find the y-intercept
2(0−6)(0+1)
f (0) = 0−2 = 6 Therefore, the y-intercept is at (0, 6).
Find the coordinates of another point.
2(3−6)(3+1)
f (3) = 3−2 = −24 Therefore, the function has a point at (3, −24). 14/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST Question: 12 Grade: 1.0 / 1.0 Find the zeros and vertical asymptote(s) of the function, if any.
2
x +2x−3
f (x) =
3x−12 If there are multiple zeros, separate the answers with semicolons (;). If there are no zeros, enter "none".
The zeros are at x = 1;-3 (50%)
Complete the equation of the vertical asymptote. If there are multiple vertical asymptotes, separate the answers with semicolons (;). For
example, if there are vertical asymptotes at x = 2 and x = 3, then enter "2;3". If there are no vertical asymptotes, enter "none".
The vertical asymptote(s) are at x = 4 (50%) Solution To find the zeros set the numerator equal to 0 and solve for x.
2
x +2x−3 = 0 (x+3)(x−1) = x x = −3 0 = 1. Thus, the zeros are x = −3 and x = 1.
To find the vertical asymptotes set the denominator equal to 0 and solve for x.
3x−12 = 0 Thus the vertical asymptote is at x = 4. 15/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 13 Grade: 1.0 / 1.0 Complete the table for the function.
2
x
+ 4x − 4 f (x) = x−3 If the answer is not defined, enter ”u”. x f (x) 1 -1/2 (17%) 3 u (17%) 5 41/2 (17%) 7 73/4 (17%) 9 113/6 (17%) 11 161/8 (17%) Solution Substitute each x -value into the function to find the corresponding f (x) -value.
2
(1) +4(1)−4
f (1) = 1−3 = 2
(3) +4(3)−4
f (3) = 3−3 = 2
(5) +4(5)−4
f (5) = 5−3 = 2
(7) +4(7)−4
f (7) = 7−3 = 2
(9) +4(9)−4
f (9) = 9−3 = 1 17
0 x f (x) 1 − 3 5 7 9 11 11−3 1
2 = undef ined 41
2
73
4
113
6 2
(11) +4(11)−4
f (11) = = − −2 = 161
8 1
2 u
41
2
73
4
113
6
161
8 16/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 14 Grade: 1.0 / 1.0 Find the vertical asymptote(s) of the function, if any.
4x+8
f (x) =
3
2
x +2x −15x Complete the equation of the vertical asymptote. If there are multiple vertical asymptotes, separate the answers with semicolons (;). For
example, if there are vertical asymptotes at x = 2 and x = 3, then enter "2;3". If there are no vertical asymptotes, enter "none".
x = -5;0;3 (100%) Solution Set the denominator equal to 0 and solve for x.
3
2
x +2x −15x = 0 x(x+5)(x−3) = 0 x = −5, x = 0, and x Thus, the vertical asymptotes are x Question: 15 = = 3 −5, x = 0, and x = 3. Grade: 1.0 / 1.0 Find the vertical asymptote(s) of the function, if any.
x+2
f (x) =
2
x −2x−8 Complete the equation of the vertical asymptote. If there are multiple vertical asymptotes, separate the answers with semicolons (;). For
example, if there are vertical asymptotes at x = 2 and x = 3, then enter "2;3". If there are no vertical asymptotes, enter "none".
x = 4 (100%) Solution Simplify.
x+2
f (x) =
2
x −2x−8
x+2
=
(x+2)(x−4)
1
=
x−4 Thus, there is only one vertical asymptote at x = 4. 17/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 16 Grade: 1.0 / 1.0 Find the vertical asymptote(s) of the function, if any.
3x−12
f (x) =
2
x +7x+10 Complete the equation of the vertical asymptote. If there are multiple vertical asymptotes, separate the answers with semicolons (;). For
example, if there are vertical asymptotes at x = 2 and x = 3, then enter "2;3". If there are no vertical asymptotes, enter "none".
x = -5;-2 (100%)
Solution Set the denominator equal to 0 and solve for x. 2
x +7x+10 = (x+5)(x+2) = x x = −5 0 0 = −2 Thus, the vertical asymptotes are x Question: 17 = −5 and x = −2. Grade: 1.0 / 1.0 Find the vertical asymptote(s) of the function, if any.
x+9
f (x) =
x−7 Complete the equation of the vertical asymptote. If there are multiple vertical asymptotes, separate the answers with semicolons (;). For
example, if there are vertical asymptotes at x = 2 and x = 3, then enter "2;3". If there are no vertical asymptotes, enter "none".
x = 7 (100%)
Solution Set the denominator equal to 0 and solve for x. Thus, the vertical asymptote is x = 7. 18/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 18 Grade: 1.0 / 1.0 Identify the graph of f (x) = 1
x+4 +4. Your response Solution Sketch the function f (x)
to x, go west.” = 1
x . Shift the graph 4 units to the left. Shift the graph 4 units up. Remember, ”If you add to y, go high, if you add 19/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 19 Grade: 1.0 / 1.0 Identify the graph of f (x) = − 1
x+ 3 . Your response Solution Sketch the basic function. Reflect the graph across the x-axis. Shift the graph 3 units to the left. Remember, ”If you add to x, go west.” 20/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 20 Grade: 1.0 / 1.0 Identify the graph of f (x) = − 1
x + 1 . Your response Solution Sketch the function. Reflect the graph across the x- axis. Shift the graph 1 unit up. Remember, ”If you add to y, you better go high.” 21/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 21 Grade: 1.0 / 1.0 Find the domain and range of the function. Your response Domain: (−∞, 1) ∪ (1, ∞)
Range: (−∞, 1) ∪ (1, ∞) Solution The domain is all values except the vertical asymptote. Therefore, the domain is any x such that x is not 1.
The range is all values except the horizontal asymptote. Therefore, the range is any y such that y is not 1. 22/23 12/2/2018 3.1 Graphing Rational Functions - PRACTICE TEST
Question: 22 Grade: 1.0 / 1.0 Choose the equations of the graph's vertical and horizontal asymptotes. Your response x = −3 and y = −1 Solution The graph approaches the vertical line x = −3. Therefore, it has a vertical asymptote at x
y = −1. Therefore, it has a horizontal asymptote at y = −1. = −3. The graph approaches the horizontal line 23/23 ...

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