Question

Which equation has a graph that lies entirely above the x-axis?

y = –(x + 7)2 + 7
y = (x – 7)2 – 7
y = (x – 7)2 + 7
y = (x – 7)2

Answer 1

`y = (x - 7)^(2) + 7`.

Explanation:

Let
`a`,
`h`, and
`k` be constants, and let
`a \\e 0`. The equation
`y = a\, (x - h)^(2) + k` represents a parabola in a plane with vertex at
`(h,\, k)`.

For example, for
`y = -(x + 7)^(2) + 7 = -(x - (-7))^(2) + 7`,
`a = (-1)`,
`h = (-7)`, and
`k = 7`.

A parabola is entirely above the
`x`-axis only if this parabola opens upwards, with the vertex
`(h,\, k)` above the
`x\!`-axis.

The parabola opens upwards if and only if the leading coefficient is positive:
`a > 0`.

For the vertex
`(h,\, k)` to be above the
`x`-axis, the
`y`-coordinate of that point,
`k`, must be strictly positive. Thus,
`k > 0`.

Among the choices:

• 
  `y = -(x + 7)^(2) + 7` does not meet the requirements. Since
  `a = (-1)`, this parabola would open downwards, not upwards as required.
• 
  `y = (x - 7)^(2) - 7` does not meet the requirements. Since
  `k = (-7)` and is negative, the vertex of this parabola would be below the
  `x`-axis.
• 
  `y = (x - 7)^(2) + 7` meet both requirements:
  `a = 1` and
  `k = 7`.
• 
  `y = (x - 7)^(2)` (for which
  `k = 0`) would touch the
  `x`-axis at its vertex.