which equation has a graph that lies entirely above the x - axis?\n$y=-(x + 7)^2+7$\n$y=(x - 7)^2-7$\n$y=(x…

which equation has a graph that lies entirely above the x - axis?\n$y=-(x + 7)^2+7$\n$y=(x - 7)^2-7$\n$y=(x - 7)^2+7$\n$y=(x - 7)^2$

which equation has a graph that lies entirely above the x - axis?\n$y=-(x + 7)^2+7$\n$y=(x - 7)^2-7$\n$y=(x - 7)^2+7$\n$y=(x - 7)^2$

Answer

Explanation:

Step1: Analyze the vertex form of a parabola

The vertex form of a parabola is (y = a(x - h)^{2}+k), where ((h,k)) is the vertex. If (a>0), the parabola opens upward; if (a < 0), the parabola opens downward.

  • For (y=-(x + 7)^{2}+7), (a=-1<0), the parabola opens downward. So it cannot lie entirely above the (x) - axis.
  • For (y=(x - 7)^{2}-7), (a = 1>0), the vertex is ((7,-7)). Since (k=-7<0), the parabola is below the (x) - axis at the vertex.
  • For (y=(x - 7)^{2}+7), (a = 1>0), the vertex is ((7,7)). Since (a>0) (opens upward) and (k = 7>0), the minimum value of (y) is (7).
  • For (y=(x - 7)^{2}), (a = 1>0), the vertex is ((7,0)). The parabola touches the (x) - axis at the vertex.

Answer:

(y=(x - 7)^{2}+7)