Question

which of the following is the graph of this square root function? f(x) = \sqrt{x + 1} + 5

Explanation:

Step1: Find the domain of the function

For the square root function \( f(x)=\sqrt{x + 1}+5 \), the expression inside the square root must be non - negative. So we set \( x + 1\geq0 \), which gives \( x\geq - 1 \). This means the graph of the function should start at \( x=-1 \) (or to the right of \( x = - 1\)).

Step2: Find the y - intercept (when \( x = 0\))

Substitute \( x = 0\) into the function \( f(x)=\sqrt{x + 1}+5 \). We get \( f(0)=\sqrt{0 + 1}+5=1 + 5=6 \)? Wait, no, wait. Wait, when \( x=-1\), \( f(-1)=\sqrt{-1 + 1}+5=\sqrt{0}+5 = 5\). Let's re - calculate the y - intercept (when \( x = 0\)): \( f(0)=\sqrt{0 + 1}+5=1 + 5 = 6\)? Wait, no, the starting point (the vertex of the square root function) for \( y=\sqrt{x - h}+k\) is at \( (h,k)\). For our function \( f(x)=\sqrt{x+1}+5=\sqrt{x-(-1)}+5 \), the vertex is at \( (-1,5)\).

Now let's analyze the three graphs:

• The first graph: The domain starts around \( x = 1\)? No, wait, looking at the x - axis, the first graph has x starting from - 1, and at \( x=-1\), the y - value is 5 (since when \( x=-1\), \( f(-1)=5\)), and as x increases, the function increases. Let's check the value at \( x = 0\): \( f(0)=\sqrt{0 + 1}+5=6\), so at \( x = 0\), \( y = 6\)? Wait, no, my mistake earlier. Wait, \( f(x)=\sqrt{x + 1}+5\). When \( x=-1\), \( f(-1)=0 + 5=5\). When \( x = 0\), \( f(0)=1 + 5=6\). When \( x = 3\), \( f(3)=\sqrt{4}+5=2 + 5=7\).

Now let's look at the three graphs:

• The first graph: The x - axis starts at - 1, and the curve starts at \( x=-1\) with \( y = 5\) (since at \( x=-1\), \( f(-1)=5\)) and as x increases, y increases. Let's check the second graph: The x - axis starts at - 8, and the curve starts at \( x=-4\) or so, which is not consistent with our domain \( x\geq - 1\). The third graph: The x - axis starts at - 2, but when \( x=-1\), let's see the y - value. Wait, the first graph: the grid, when x=-1, the y - coordinate is 5 (from the graph's grid, the first graph has y - axis with 5 at the level where x=-1 starts). The second graph has a lower starting y - value, and the third graph has a starting y - value around 5.5? No, wait, let's re - evaluate.

Wait, the standard square root function \( y=\sqrt{x}\) has a domain \( x\geq0\) and starts at (0,0). The function \( y=\sqrt{x + 1}\) is a horizontal shift of \( y=\sqrt{x}\) to the left by 1 unit, so its domain is \( x\geq - 1\) and it starts at (- 1,0). Then the function \( y=\sqrt{x + 1}+5\) is a vertical shift of \( y=\sqrt{x + 1}\) up by 5 units, so it starts at (- 1,5) and increases as x increases.

Now let's check the three graphs:

• First graph: The x - axis starts at - 1, and the curve starts at \( x=-1\) with \( y = 5\) (matches the starting point (- 1,5)) and as x increases, the curve goes up (which is the behavior of a square root function, increasing at a decreasing rate).

• Second graph: The curve starts at \( x=-4\) or so, which is outside the domain \( x\geq - 1\) of our function, so it's incorrect.

• Third graph: When \( x=-1\), let's see the y - value. The third graph's curve starts at \( x=-1\) with \( y\approx5.5\) or so, but our function at \( x=-1\) should be \( y = 5\). Also, the first graph's starting point is at \( x=-1,y = 5\), which matches the vertex of our function \( f(x)=\sqrt{x + 1}+5\).

Answer:

The first graph (the left - most graph) is the graph of the function \( f(x)=\sqrt{x + 1}+5\).