Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Let $a$, $h$, and $k$ be constants, and let $a \ne 0$. 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.