This question looks like exercises in transformations. When you have subtraction inside the parenthesis, the graph moves that many units to the right. So, for the first one, f(x-3), the entire graph moves 3 units to the right. So, all the intercepts (x intercepts, where the graph cuts the x axis) move 3 units to the right. So, the x intercepts become (1,0), (7,0), and (10,0).
For the second one, the graph gets twice as...
This question looks like exercises in transformations. When you have subtraction inside the parenthesis, the graph moves that many units to the right. So, for the first one, f(x-3), the entire graph moves 3 units to the right. So, all the intercepts (x intercepts, where the graph cuts the x axis) move 3 units to the right. So, the x intercepts become (1,0), (7,0), and (10,0).
For the second one, the graph gets twice as tall, i.e grows in height. But, this means the x intercepts would still stay the same. So, the x intercepts are still (-2,0), (4,0), and (7,0).
On the third one, the graph flips around the x axis, from top to bottom, bottom to top. However, this also means the intercepts would stay the same. So, the x intercepts are still (-2,0), (4,0), and (7,0).
For the 4th one, the graph would flip around the y axis. So, then, the intercepts get "mirrored" around the y axis. For example, (-2,0) becomes (2,0). So, the x intercepts here would become (2,0), (-4,0), and (-7,0)
The 5th one has addition outside the parenthesis. As opposed to the first one (subtraction on the inside of the parenthesis), the entire graph now moves up that many units. So would the maximum turning point, or the highest point. So, (2,5) would become (2,8)
In the last one, the graph again flips about the x axis, or upside down. So, all the maximum points becomes minimum points, and all the minimum points become maximum points. So, for this one, the maximum point would be (5,1).
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