The first graph (problem 4) represents an absolute value function because of its V-shape. The graph passes through points (1, 1) and (-1, 1).
Since the slope of the lines is 1, and the graph opens upwards, the equation will have a positive coefficient. Thus, the equation that corresponds to graph 4 is y = |x|.
The second graph (problem 5) also represents an absolute value function. The graph passes through the points (1, -1) and (-1,-1).
Since the slope of the lines is -1 and the graph opens downwards, the function will have a negative coefficient. The vertex of the graph is at (0, 0), thus there is no vertical shift. The equation that corresponds to graph 5 is y = -|x|.
The third graph (problem 6) represents a quadratic function due to its U shape. It opens upward meaning the squared term will have a positive coefficient. The vertex of the parabola is at (1, 0) indicating a horizontal shift.
Since the vertex is at (1,0), and the graph passes through the point (2, 1), we can find the equation. The general form is y = a(x-h)^2 + k where (h, k) is the vertex. In this case the vertex is (1, 0), so the equation will be of the form y=a(x-1)^2. Substituting (2, 1), we have 1 = a(2-1)^2 = a. So, a = 1. Therefore, the equation is y = (x-1)^2 = x^2 - 2x + 1
- y = |x|
- y = -|x|
- y = (x-1)² or y = x² - 2x + 1
Answered
