Question

identify the equation for the graph.
$y - 2 = \sqrt{x}$
$y + 2 = \sqrt{x}$
$y = \sqrt{x} + 2$

Explanation:

Step1: Analyze the graph's starting point

The graph starts at \((0, -2)\) (since it touches the y - axis at \(x = 0\) and \(y=-2\)). We can test this point in each equation.

Step2: Test the point \((0, -2)\) in the first equation \(y - 2=\sqrt{x}\)

Substitute \(x = 0\) and \(y=-2\) into \(y - 2=\sqrt{x}\):
Left - hand side: \(-2-2=-4\)
Right - hand side: \(\sqrt{0} = 0\)
Since \(-4
eq0\), this equation is not correct.

Step3: Test the point \((0, -2)\) in the second equation \(y + 2=\sqrt{x}\)

Substitute \(x = 0\) and \(y=-2\) into \(y + 2=\sqrt{x}\):
Left - hand side: \(-2 + 2=0\)
Right - hand side: \(\sqrt{0}=0\)
This equation holds for the point \((0, -2)\). Let's also check the shape. The square - root function \(y=\sqrt{x}\) has a domain \(x\geq0\) and a range \(y\geq0\). The equation \(y + 2=\sqrt{x}\) can be rewritten as \(y=\sqrt{x}-2\), which shifts the graph of \(y = \sqrt{x}\) down by 2 units, which matches the starting point \((0,-2)\) and the general shape of a square - root function.

Step4: Test the point \((0, -2)\) in the third equation \(y=\sqrt{x + 2}\)

Substitute \(x = 0\) and \(y=-2\) into \(y=\sqrt{x + 2}\):
Left - hand side: \(-2\)
Right - hand side: \(\sqrt{0 + 2}=\sqrt{2}\approx1.414\)
Since \(-2
eq\sqrt{2}\), this equation is not correct.

Answer:

\(y + 2=\sqrt{x}\)