What is the domain (in interval notation) of the following functions?
1 answer:
You might be interested in
Part A
The graph is shown below as an attached image.
The diagram shows a straight line that goes through the two points (0,-3) and (1, -5)
I'm using GeoGebra to graph the line.
side note: (0, -3) is the y intercept which is where the graph crosses the y axis.
==================================================
Part B
Answer is choice 2
The graph can be written in the form y = mx+b, so it is linear
In this case, m = -2 is the slope and b = -3 is the y intercept
We can write the slope as m = -2 = -2/1. This tells us that we can move down 2 units and then over to the right 1 units to get from point to point. This process of "down 2, over to the right 1" happens when moving from point A to point B in the diagram below.
The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)
<h3>How to match the vertices of the polygon?</h3>The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.
The vertices of polygon ABCD are given as:
A = (4, 6)
B = (5, 3)
C = (4, 2)
D = (1, 2)
The rule of rotation about point C is:
(x,y) = (a + b - y, x + b - a)
Where:
(a, b) = (4, 2) --- the point of rotation.
So, we have:
(x,y) = (4 + 2 - y, x + 4 - 2)
(x,y) = (6 - y, x + 2)
The above means that:
A' = (6 - 6, 4 + 2) = (0,6)
B' = (6 - 3, 5 + 2) = (3,7)
C' = (6 - 2, 4 + 2) = (4,6)
D' = (6 - 2, 1 + 2) = (4,3)
Hence, the image of the rotation and their vertices (i.e. coordinates) are:
A' = (0,6)
B' = (3,7)
C' = (4,6)
D' = (4,3)
Read more about rotation at:
#SPJ1
