Question

identify the graph for the equation.
$y - 3 = \sqrt{x - 2}$
three graphs are shown, each with a coordinate grid and a blue curve. the first graph has axes from -8 to 8, the second from -8 to 8, and the third from -2 to 8 on y and -2 to 6 on x.

Explanation:

Step1: Analyze the domain of the function

The expression under the square root must be non - negative. So, for \(y - 3=\sqrt{x - 2}\), we have \(x-2\geq0\), which implies \(x\geq2\). This means the graph of the function should start at \(x = 2\) or to the right of \(x = 2\).

Step2: Analyze the range of the function

Since the square root function \(\sqrt{x - 2}\geq0\) (because the square root of a non - negative number is non - negative), then \(y-3=\sqrt{x - 2}\geq0\), so \(y\geq3\).

Step3: Analyze the vertex (starting point)

When \(x = 2\), \(\sqrt{2 - 2}=0\), then \(y-3 = 0\), so \(y=3\). So the graph should start at the point \((2,3)\).

Now let's analyze the three graphs:

• The first graph: The starting point seems to be at \(x>2\) but \(y<3\), so it does not satisfy the range condition.
• The second graph: The starting point has \(x<2\) (since \(x\) values are negative or less than 2), which violates the domain condition \(x\geq2\).
• The third graph: The starting point is around \(x = 2\) and \(y = 3\) (or close to it), and as \(x\) increases, \(y\) increases (since \(y=3+\sqrt{x - 2}\) and as \(x\) increases, \(\sqrt{x - 2}\) increases, so \(y\) increases). It also satisfies the domain \(x\geq2\) and range \(y\geq3\) conditions.

Answer:

The third graph (the one with the y - axis labeled from - 2 to 8 and x - axis from - 2 to 6, starting near (2,3) and increasing)