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Which graph represents x > (y – 2)2 – 4? On a coordinate plane, a parabola opens up and goes through (0, 0), has a vertex at (2, negative 4), and goes through (4, 0). Everything outside of the parabola is shaded. On a coordinate plane, a parabola opens up and goes through (0, 0), has a vertex at (2, negative 4), and goes through (4, 0). Everything inside of the parabola is shaded. On a coordinate plane, a parabola opens to the right and goes through (0, 4), has a vertex at (negative 4, 2), and goes through (0, 0). Everything inside of the parabola is shaded. On a coordinate plane, a parabola opens to the right and goes through (0, 4), has a vertex at (negative 4, 2), and goes through (0, 0). Everything outside of the parabola is shaded.

User ChrisShick
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1 Answer

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The given inequality is x > (y – 2)2 – 4. To graph this inequality, we first need to plot the vertex of the parabola, which is (2, -4). This point is included in the shaded region because the inequality is "greater than" and not "greater than or equal to". Next, we need to determine whether the parabola opens upward or downward. Since the coefficient of the squared term is positive, the parabola opens upward.

Now, we need to decide which side of the parabola to shade. To do this, we can choose a test point that is not on the parabola and substitute its coordinates into the inequality. For example, the point (0,0) is a convenient test point. Substituting x=0 and y=0 into the inequality, we get:

0 > (0 – 2)2 – 4

0 > 0

Since 0 is not greater than 0, the point (0,0) is not in the shaded region. Therefore, we need to shade the side of the parabola that does not contain the point (0,0).

Among the given options, only the first one matches the graph described above. Therefore, the answer is: the graph that represents x > (y – 2)2 – 4 is the one where a parabola opens up and goes through (0, 0), has a vertex at (2, -4), and everything outside of the parabola is shaded.

User Montoya
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