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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

1 Answer

11 votes

Answer:


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.
User EMBarbosa
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