Use addition to solve the linear system of equations. Include all of your work in your final answer.
1 answer:
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All the points that are 6 units from (-1, 1) are those on the circle
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
(x+1)^2 +(y-1)^2 = 36
For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35
The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).
Answer:
The last listed functional expression:
Step-by-step explanation:
It is important to notice that the two linear expressions that render such graph are parallel lines (same slope), and that the one valid for the left part of the domain, crosses the y-axis at the point (0,2), that is y = 2 when x = 0. On the other hand, if you prolong the line that describes the right hand side of the domain, that line will cross the y axis at a lower position than the previous one (0,1), that is y=1 when x = 0. This info gives us what the y-intercepts of the equations should be (the constant number that adds to the term in x in the equations: in the left section of the graph, the equation should have "x+2", while for the right section of the graph, the equation should have x+1.
It is also important to understand that the "solid" dot that is located in the region where the domain changes, (x=2) belongs to the domain on the right hand side of the graph, So, we are looking for a function definition that contains for the function, for the domain:
.
Such definition is the one given last (bottom right) in your answer options.