Answer 1: P1: (X1, Y1) =(0,0) and P2: (X2, Y2)=(4,6), slope=3/2
In this part we can take the point of the begining and the end of the red line. Taking into account each point has an X component and a Y component:
P1: (X1, Y1) =(0,0) and P2: (X2, Y2)=(4,6)
With this two points we can use the equation of the slope [tex]m[/tex]:
[tex]m=\frac{Y2-Y1}{X2-X1}[/tex]
[tex]m=\frac{6-0}{4-0}=\frac{3}{2}[/tex] This is the slope of the red line
Now we have to write the equation of the line in the form [tex]y=mx+b[/tex]
Where [tex]b[/tex] is the intersection point with the y-axis and can be calculated using one of the points (we will use (4,6)) and the slope ([tex]m=\frac{3}{2}[/tex]):
[tex]6=\frac{3}{2}(4)+b[/tex]
[tex]b=0[/tex]
Hence: [tex]y=\frac{3}{2}x[/tex] This is the equation of the red line
Answer 2: P1: (X1, Y1) =(4,6) and P2: (X2, Y2)=(6,2), slope=-2
Here we will approach the problem similarly as we did in part 1:
Two end points of blue line: P1: (X1, Y1) =(4,6) and P2: (X2, Y2)=(6,2)
Calculating the slope:
[tex]m=\frac{Y2-Y1}{X2-X1}[/tex]
[tex]m=\frac{2-6}{6-4}=-2[/tex] This is the slope of the blue line (note the negative sign indicates it is a decline line)
Calculating the intersection point (we will use (4,6)):
[tex]y=mx+b[/tex]
[tex]6=-2(4)+b[/tex]
[tex]b=14[/tex]
Hence, the linear equation of the decline blue line is:
[tex]y=-2x+14[/tex]
Answer 3: P1: (X1, Y1) =(6,2) and P2: (X2, Y2)=(10,5), slope=3/4
For the green line we have:
P1: (X1, Y1) =(6,2) and P2: (X2, Y2)=(10,5)
For the slope:
[tex]m=\frac{Y2-Y1}{X2-X1}[/tex]
[tex]m=\frac{5-2}{10-6}=\frac{3}{4}[/tex] This is the slope of the green line
Answer 4: Blue line is the steepest of the three.
Now we have to compare the slope of each of the three lines:
Red line: [tex]\frac{3}{2}=1.5[/tex]
Blue line: [tex]-2[/tex]
Green line: [tex]\frac{3}{4}=0.75[/tex]
How mathematically can we tell it is the steepest?
As we can see blue line has the greatest slope (remember the negative sign only indicates the direction of the slope). This means the Blue line is the steepest of the three.
Answer 5: The orange track is not a function
How do we know it?
Graphically, a function is identified by drawing a vertical line on it, if this line cuts to the plotted line at more than one point, this is not a function.
In this sense, if we draw a vertical line in the last region of the graph, this line will cut it in more than one point. Therefore it is not a function.
Answer 6: [tex]0 \leq x \leq 12 [/tex] or [0,12]
A function is represented graphically by a correspondence between the elements of the domain set and those of the range set.
What does this mean?
The Domain of a function is the set of all the real values the variable X can take and that can be seen in the graph.
For example, in this case the roller coaster is defined for all values of X that are between 0 and 12 (both included)
Answer 7: [tex]0 \leq y \leq 8 [/tex] or [0,8]
The Range of a function refers to the values taken by the "Y" function (dependent variable), since its value depends on the value we give to "X".
Graphically, the way to determine the Range is to graph the function and see the values that "Y" takes from the bottom up. That is, the values that have the variable “y” for certain values of x.
For example, in this graph of the roller coaster the range is defined for all the values of Y between 0 and 8 (both included).
