Week 5: Student Response To Discussion 1

ThursdayAug 17 at 4:06pm

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The equations I was given are x + 3 = |y| and y = 4|x|, which I solved and broke down to the best of my abilities.

x + 3 = |y| requires a handful of x-values to be input in order to find various y-values which in turn gives us a multitude of number sets, which we can plot of an XY graph. Using -5, -4, -3, -2, and 1, we plug each into the equation (which requires the addition of these numbers with positive three and solving the solving the absolute value of the outcome) and we are able to get the y-values of 2, 1, 0, 1, and 2, which create the following sets: (-5, 2) (-4, 1) (-3, 0) (-2, 1) (-1, 2). When graphed, the vertex and intercept are at (-3, 0) and the shape is a V which opens upwards. It is a function because I used the vertical line test, requiring me to drop a number of vertical lines into my graph to which these lines never entered the parabola twice. The domain for this graph is (-∞,∞) and the range is [-3, ∞). The ordered sets of numbers are in relation to each other in that they create a foreseeable and consistent pattern in there states.

y = 4|x| requires us to plug another handful of possible x-values into the equation to sort out some y-values. Using -2, -1, 0, 1, and 2, we create y-values of 8, 4, 0, 4, and 8 (after multiplying each by the absolute value of the plugged values), which give us the following sets: (-2, 8) (-1, 4) (0, 0) (1, 4) (2, 8). Plotting these points gives us a V-shaped parabola with a vertex and intercept of (0, 0). It is a function in that the vertical line test proved otherwise, the domain is (-∞,∞) and the range is [0, ∞).

For y = 4|x|, if we were to move the graph up three units and then to the left four units, this would be a transformation. The equations would then look like this: y = 4 +3|x+4| or y = 7|x+4|.