If you were given the equation f ( x ) = 3x⁵+6x, what behavior would the ends of the graph have?
1 answer:
The function given in the problem is a Polynomial Function of degree 5. These functions have the following form: f(x)=ax^5+bx^4+c^3+d^2+ex+f. They have an R domain and their graphs are continuous curves.
You can know the behavior of the function f(x)=3x⁵+6x by given values to the variable x and plot each point obtained. For example:
If x=-1⇒ f(-1)=3(-1)⁵+6(-1)=9
If x=0⇒ f(0)==3(0)⁵+6(0)=0
If x=1⇒ f(1)=3(1)⁵+6(1)=9
Therefore, as you can see in the graph attached, <span>the correct answer is:
c. The left end will start down at a large negative and the right end will go upwards.</span>
You can know the behavior of the function f(x)=3x⁵+6x by given values to the variable x and plot each point obtained. For example:
If x=-1⇒ f(-1)=3(-1)⁵+6(-1)=9
If x=0⇒ f(0)==3(0)⁵+6(0)=0
If x=1⇒ f(1)=3(1)⁵+6(1)=9
Therefore, as you can see in the graph attached, <span>the correct answer is:
c. The left end will start down at a large negative and the right end will go upwards.</span>
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Answer:
Step-by-step explanation:
Hello!
Data
Year/Firefighters
2000: 20
2001: 18
2002: 23
2003: 30
2004: 20
2005: 12
2006: 24
2007: 9
2008: 25
2009: 15
2010: 8
2011: 11
2012: 15
2013: 34
<u>Common measures of variation</u>
<u>Variance</u>
S²=
∑X= 20 + 18 + 23 + 30 + 20 + 12 + 24 + 9 + 25 + 15 + 8 + 11 + 15 + 34
∑X= 264
∑X²= 20² + 18² + 23² + 30² + 20² + 12² + 24² + 9² + 25² + 15² + 8² + 11² + 15² + 34²
∑X²= 5770
n= 14
S²=
S²= 60.90
<u>Standard deviation</u>
The standard deviation is the square root of the variance
S= √S²= √60.90= 7.80
<u>Coefficient of variation</u>
Is a relative standard deviation, it is defined as the division of the standard deviation (S) by the mean (X[bar)]. It has not units and is usually expressed in percentage. It shows the variability in relation to the mean.
C.V.= *100 =
*100= 41.36%
The mean or average is a measurement of position and gives you an idea of the central value of the distribution of the data.
X[bar]= ∑X/n= 264/14= 18.86
<u>Range</u>
In the interval between the max and min values. It allows you to have an idea of the dispersion of the values.
R= Xmax - Xmin= 34 - 8= 26
Xmax= 34
Xmin= 8
<u>Inter Quartile Range</u>
Is the difference between Quantile 1 (C₁) and Quantile 3 (C₃). It shows the 50% mid values of the sample.
IQR= C₃ - C₁= 23.5 - 11.5 = 12
Quantile 1 (C₁) is the value that leaves 25% of the sample below and 75% of the sample above.
Pos1: n/4= 14/4= 3.5
The first quantile is the number between position 3 and position 4
8, 9, 11, 12, 15, 15, 18, 20, 20, 23, 24, 25, 30, 34
C₁= (11+12)/2= 11.5
Quantile 3 (C₃) is the value that leaves 75% of the sample below and 25% of the sample above.
Pos3: n*3/4= 14*3/4= 10.5
The third Quantile is between position 10 and 11
8, 9, 11, 12, 15, 15, 18, 20, 20, 23, 24, 25, 30, 34
C₃= (23+24)/2= 23.5
The measurements of variation don't allow you to know the origin of the data, that it is from consecutive years.
I hope it helps!