Answer:
D
Step-by-step explanation:
Let's derive the equations for the two graphs;
For the graph towards the right,
The slope = y2-y1/x2-x1; so we choose two corresponding points (x1,y1); (x2,y2)
Let's choose points;
(2,1) and (3,2)
The slope = 2-1/ 3-2 = 1/1 = 1
From the general equation of a line
y = mx + c = 1× x + (-1) = x-1
y = x-1
Similarly for that towards the left we have;
The slope = y2-y1/x2-x1; so we choose two corresponding points (x1,y1); (x2,y2)
Let's choose points;
(2,1) and (3,2)
The slope = 2-1/ 3-2 = 1/1 = 1
From the general equation of a line
y = mx + c = 1× x + (-1) = x-1
y = x-1
Similarly for that towards the left we have;
The slope = y2-y1/x2-x1; so we choose two corresponding points (x1,y1); (x2,y2)
Let's choose points;
(-7,0) and (-4,-3)
Slope = -3-0/ -4-(-7) = -3/ 3 = -1
From the general equation of a line
y = mx + c = -1× x + c = -x + c
Since c is not obvious we can find the equation by picking just a point along the graph along side an arbitrary (x,y) point hence we have for slope;
If we pick (-7,0)
y-0 / x-(-7) = -1
y / x+7 = -1
y = -x -7
The equations of the graphed lines are;
y=x-1
y=-x-7
Look at option D you we see it's the same as the equation of the graph towards the right y=x-1
For the records
y=[x+3]-4 = x+3-4 = x-1
........a+b <span>≥ 0</span>