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"Please see the graphs in the attachments.
For A. Notice that the form of the functions is just .
A1. First graph the original x^2. Remember that a quadratic function looks like a parabola. The original points upwards, and has its vertex at (0,0). An easy way to graph this is to plot the points (0,0), (1,1), and (1,1). Just hit those three as you graph your parabola.
A2 and A3. Note that when the coefficient gets bigger, the graph ""squishes"" closer to the yaxis. This makes sense if you think about it since, when y was x^2, at x =1, y =1. But when y = 2x^2, at x =1, y = 2, and so on. So the y value gets higher faster. If this were linear, we would say that it has a steeper slope. To graph y = 2x^2, we can also use points (0,0), (1,2), and (1, 2).
A4 and A5. Here, you can see that the graph expands. Similar to the explanation for A2 and A3, here, the functions increase slower than the original x^2. This is since the coefficient for x^2 is 1, then, if 0 < a < 1, the graph will stretch. The closer it is to 0, the ""fatter"" it will be.
A6 and A7. Here, a < 0. When you take the negative of the function, just reflect it across the xaxis. So x^2 is just the mirror image of x^2, and 2x^2 is the mirror image of 2x^2.
For B. The form of the functions is just .
B1 is just the same x^2 as in part A.
B2 and B5. When h > 0, the function moves to the right by h units. So in B2, instead of the vertex being at (0,0), it moves 2 units to the right to become (2,0). Similarly for B5, the vertex moves 1 unit to (1,0).
B3 and B4. When h < 0, the function moves to the left by h units. In the same way, in B3, the vertex moves from the original (0,0) to (2,0). (1,1) and (1, 1) from the original also move 2 units to the left, so they become (1,1) and (3,1) respectively.
Here are some more information:
brainly.ph/question/18908
brainly.ph/question/425659
brainly.ph/question/1858171
Hope this helps!"
