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# Whose the answer of math 9 module 2 to the left to the right put me up put me down letter d?

• Réponse publiée par: cleik

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 y-axis. 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 x-axis. 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.