1. Problem # 14: F(x) = |x – 2|
  1. Problem # 48: x = 1 – y2
  • There are many ways to go about solving math problems. For this assignment you will be required to do some work that will not be included in the discussion. First, you need to graph your functions so you can clearly describe the graphs in your discussion. Your graph itself is not required in your post, although the discussion of the graph is required. Make sure you have at least five points for each equation to graph. Show all math work for finding the points.
  • Specifically mention any key points on the graphs, including intercepts, vertex, or start/end points. (Points with decimal values need not be listed, as they might be found in a square root function. Stick to integer value points.)
  • Discuss the general shape and location of each of your graphs.
  • State the domain and range for each of your equations. Write them in interval notation.
  • State whether each of the equations is a function or not giving your reasons for the answer.
  • Select one of your graphs and assume it has been shifted three units upward and four units to the left. Discuss how this transformation affects the equation by rewriting the equation to incorporate those numbers.
  • Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing the thought behind your math work.):
  • Function
  • Relation
  • Domain
  • Range
  • Transformation

(1) F(x) = |x - 2|
  • Values selected for x are -4, -3, -2, -1, 0, 1, 2, 3, 4
  • These values are plugged into the given equation and F(x) is calculated.
    x = -4, F(x) = |-4 - 2| = |-6| = 6
    x = -3, F(x) = |-3 - 2| = |-5| = 5
    x = -2, F(x) = |-2 - 2| = |-4| = 4
    x = -1, F(x) = |-1 - 2| = |-3| = 3
    x = 0, F(x) = |0 - 2| = |-2| = 2
    x = 1, F(x) = |1 - 2| = |-1| = 1
    x = 2, F(x) = |2 - 2| = |0| = 0
    x = 3, F(x) = |3 - 2| = |1| = 1
    x = 4, F(x) = |4 - 2| = |2| = 2
  • Here is the table:
x / F(x) = |x - 2|
-4 / 6
-3 / 5
-2 / 4
-1 / 3
0 / 2
1 / 1
2 / 0
3 / 1
4 / 2
  • Graph is shown below:

  • x- intercepts: Points where the graph crosses the x- axis. It is (2, 0)
  • y- intercepts: Points where the graph crosses the y- axis. It is (0, 2)
  • Since this is the graph of an absolute value function, it is shaped like a V. The graph is located only in the I and II quadrants.
  • Since we can assign any value for x, the domain is the set of all real numbers (-, )
  • Since the graph is only in the I and II quadrants F(x) is never negative. The range is [0, )
  • For every value of x there is one and only one value of F(x). Hence the graph is that of a function. A vertical line (if drawn on the graph) will cross the graph only once. This confirms that the graph is that of a function.
  • If the graph is shifted 3 units upward and 4 units to the left, the new equation will be
    G(x) = |(x + 4) - 2| + 3, that is G(x) = |x + 2| + 3
  • The new table and transformed graph are shown below:
x / G(x) = |x + 2| + 3
-4 / 5
-3 / 4
-2 / 3
-1 / 4
0 / 5
1 / 6
2 / 7
3 / 8
4 / 9

(2) x = 1 - y2 y2 = 1 - x  y =
  • Values selected for x are -24, -15, -8, -3, 0, 1 so that the y values come as integers only and not as decimals
  • These values are plugged into the given equation and F(x) is calculated.
    x = -24, y = = √25 = 5
    x = -15, y = = √16 = 4
    x = -8, y = = √9 = 3
    x = -3, y = = √4 = 2
    x = 0, y = = √1 = 1
    x = 1, y = = √0 = 0
  • Here is the table:
x / y = √(1 - x)
-24 / 5
-15 / 4
-8 / 3
-3 / 2
0 / 1
1 / 0
  • Graph is shown below:

  • x- intercepts: Points where the graph crosses the x- axis. It is (1, 0)
  • y- intercepts: Points where the graph crosses the y- axis. It is (0, 1)
  • Since this is the graph of a square root function, it is shaped like a parabola with its vertex at (1, 0) and opening to the left. The graph is located only in the I and II quadrants.
  • The function is defined only when the value under the square root is non-negative. So we can assign values for x such that x is 1 or lesser. The domain is (-, 1].
  • Since the graph is only in the I and II quadrants y is never negative. The range is [0, )
  • For every value of x there is one and only one value of y. Hence the graph is that of a function. A vertical line (if drawn on the graph) will cross the graph only once. This confirms that the graph is that of a function.
  • If the graph is shifted 3 units upward and 4 units to the left, the new equation will be
    y = + 3,that is Y = + 3,
  • The new table and transformed graph are shown below:
x / Y = √[-(x + 3)] + 3
-28 / 8
-19 / 7
-12 / 6
-7 / 5
-4 / 4
-3 / 3