Algebra Homework: Analyzing Functions, Lines, and Regression Analysis

This document presents a comprehensive solution to an algebra homework assignment. It addresses various topics including piecewise functions, graphing functions, determining whether functions are even or odd, finding x and y intercepts, and analyzing the domain and range of functions. The solution also covers linear equations, including finding the equation of a line given points, and determining the equation of a perpendicular line. Furthermore, the assignment explores real-world applications through a word problem involving a phone bill calculation. Finally, it delves into regression analysis, including the calculation of the least squares regression line, correlation coefficient, and interpretation of the slope, demonstrating a strong understanding of algebraic concepts and their applications. This assignment is contributed by a student to be published on the website Desklib. Desklib is a platform which provides all the necessary AI based study tools for students.

ALGEBRA
STUDENT NAME/ID
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Question 1
Given piecewise function
f ( x )={
(a) f ( 0 ) =?
Here,
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2 x +3 for x ≤0 would be taken into account
Now,
f ( 0 )= (3
2 )∗0+ 3=3
(b)f ( 2 ) =?
Here,
( x−1 ) 2 +2 for 0< x ≤ 2 would be taken into account
Now,
f ( 2 ) = ( 2−1 )2 +2=3
Question 2
The correct graph would be Graph A. This is because f(x+1) would involve shifting of the
whole graph of f(x) one unit to the right and -1 would imply a downward vertical shift for every
value of the graph. Both these attributes are captured in graph A.
Question 3
f ( x )=3 x2−x3
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A function would be even when f (−x)=f ( x) for all x ∈ R
A function would be odd when f ( −x ) =−f ( x) for all x ∈ R
Now,
f ( x )=3 x2−x3
f (−x ) =3 (−x )2 − (−x )3
f ( −x ) =3 x2 + x3
It can be seen from the above that f (−x ) ≠ f ( x )and hence, function is not even function.
Further, f ( −x ) ≠−f ( x ) and hence, function is not odd function.
Therefore, it can be concluded that given function f ( x )=3 x2−x3is neither an even nor an odd
function.
Question 4
a) x axis – NO, y axis- No, Origin – Yes
b) x axis – NO, y axis- Yes, Origin – NO
Question 5
a) The given function is increasing on (-∞,0] U [3.5]
b) Domain: (--∞,,5] Range: (--∞,,4]
Question 6
Let
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A(3 ,−5)
B(−1 , 7)
(a) Equation of line passing through point A and B
Equation of line y = mx + c
Slopeof the equation m=7−(−5)
−1−3 =−3
Now,
−5= ( −3 )∗3+c
c=4
y=mx+c
y=−3 x + 4
Hence, Equation of line passing through point A and B, y= -3x+4.
(b) Equation of line which is perpendicular to y = -3x+4 and passes through (1, 7)
Line which is perpendicular to y = -3x+4 would have a reciprocal slope which means, slope of
line = 1/3
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Now,
y=mx+c
y=1 /3 x +c
Passes through (1, 7)
7= ( 1/ 3 ) +c
c=20/3
Hence,
y= x
3 + 20
3
3y = x+20
Question 7
Company charged $25 monthly fee for 100 minutes
Additional fee $0.07 per minutes for every minute higher than 100 minutes
(a) Let B(x) is the monthly phone bill and x minutes that are over than 100 minutes.
Total bill amount would be computed based on the equation derived as shown below.
B ( x ) =$ 25+ ( $ 0.07∗x )
B ( x ) =0.07 x+ 25
(b) Bill amount when the total minute used in month is 650 minutes.
Total minutes that are over than 100 minutes (x) = 650-100 = 550 minutes
Hence,
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B ( x ) =(0.07∗550)+25
B ( x ) =$ 63.5
(c) Number of minutes used if the monthly bill of the subscriber is $84
The monthly phone bill B(x) = $84
Hence,
B ( x ) =(0.07 x )+ 25
84=0.07 x +25
x=842.85 minutes
Hence, total minutes used would be = 100 + 842.85 = 942.85 minutes
Question 8
The x and y intercepts of the function.
f ( x )=2|x−3 |−6
 x intercept is a point on the graph where y = 0,
Put y = f(x) = 0
0=2|x−3|−6
2 |x−3|=6
|x−3|=3
Now, |f ( x )|=a
f ( x )=−a∨f ( x )=a
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x−3=−3 , x=0
x−3=3 , x =6
Hence, x intercept (0, 0) and (6, 0)
 y intercept is a point on the graph where x =0
Put x = 0
y=2|0−3|−6
y=(2∗(3))−6
y=0
Hence, y intercept (0, 0)
Question 9
(a) Least squares regression line
Temperature = Independent variable (X)
Number of chirps per second = Dependent variable (Y)
y=0.21 x +0.56
Number of chirps per second = (0.21 * Temperature) + 0.56
(b) Correlation coefficient
The value of correlation coefficient R = 0.91
The correlation coefficient comes out to be positive which indicates the positive association
between the variables. Further, the value is close to 1 which is theoretically shows perfect strong
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correlation between the variables and hence, it can be said that strength of correlation between
temperature and number of chirps per second is strong.
(c) Slope of regression line = 0.21
The slope indicates that as the temperature tends to change by 1 unit, the corresponding number
of chirps per second would change by 0.21 units in the same direction.
(d) Average number of chirps per second =?
Temperature = 600F
Number of chirps per second = (0.21 * Temperature) + 0.56
Number of chirps per second = (0.21 * 60) + 0.56 = 12.94
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