Answer :
To graph the function [tex]\( f(x) = \sqrt{x + 3} - 2 \)[/tex], follow these steps to understand its characteristics and behavior:
1. Identify the domain:
The function [tex]\( f(x) = \sqrt{x + 3} - 2 \)[/tex] is defined when the expression inside the square root is non-negative, i.e., [tex]\( x + 3 \geq 0 \)[/tex]. This implies:
[tex]\[ x \geq -3 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is [tex]\( x \in [-3, \infty) \)[/tex].
2. Determine the range:
The output of the square root function [tex]\( \sqrt{x + 3} \)[/tex] is non-negative (i.e., [tex]\( \geq 0 \)[/tex]) for [tex]\( x \geq -3 \)[/tex]. The smallest value of [tex]\( \sqrt{x + 3} \)[/tex] occurs when [tex]\( x = -3 \)[/tex] and is zero. Subtracting 2 shifts the entire function down by 2 units. Thus, [tex]\( \sqrt{x+3} - 2 \)[/tex] starts at [tex]\(-2\)[/tex] when [tex]\( x = -3 \)[/tex] and increases without bound as [tex]\( x \)[/tex] increases. Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\( y \in [-2, \infty) \)[/tex].
3. Find the x-intercept:
To find the x-intercept, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \sqrt{x + 3} - 2 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x + 3} = 2 \quad \Rightarrow \quad x + 3 = 4 \quad \Rightarrow \quad x = 1 \][/tex]
Thus, the x-intercept occurs at [tex]\( (1, 0) \)[/tex].
4. Find the y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{0 + 3} - 2 = \sqrt{3} - 2 \][/tex]
Since [tex]\( \sqrt{3} \approx 1.732 \)[/tex], we have:
[tex]\[ f(0) \approx 1.732 - 2 = -0.268 \][/tex]
Thus, the y-intercept is approximately [tex]\( (0, -0.268) \)[/tex].
5. Plot additional points:
Calculate a few more points within the domain to understand the function's behavior:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \sqrt{0} - 2 = -2 \][/tex]
- When [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \sqrt{1} - 2 = 1 - 2 = -1 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \sqrt{4} - 2 = 2 - 2 = 0 \][/tex]
- When [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{10} - 2 \approx 3.162 - 2 = 1.162 \][/tex]
6. Sketch the graph:
Plot the points [tex]\((-3, -2), (1, 0), (0, -0.268), (-2, -1), (7, 1.162)\)[/tex] and draw a curve passing through these points keeping in mind the domain [tex]\( x \geq -3 \)[/tex]. The graph starts at the point [tex]\((-3, -2)\)[/tex] and rises to the right.
The graph of [tex]\( f(x) = \sqrt{x + 3} - 2 \)[/tex] is thus a transformed square root function that has been shifted 3 units to the left (because of [tex]\( x + 3 \)[/tex] under the square root) and 2 units down (because of the [tex]\(-2\)[/tex]). It opens to the right from the starting point [tex]\((-3, -2)\)[/tex].
So, the correct graph is the one that starts at [tex]\((-3, -2)\)[/tex], passes through [tex]\( (1, 0) \)[/tex], and continues to rise slowly as [tex]\( x \)[/tex] increases, following the standard shape of a square root function adjusted by the transformations described above.
1. Identify the domain:
The function [tex]\( f(x) = \sqrt{x + 3} - 2 \)[/tex] is defined when the expression inside the square root is non-negative, i.e., [tex]\( x + 3 \geq 0 \)[/tex]. This implies:
[tex]\[ x \geq -3 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is [tex]\( x \in [-3, \infty) \)[/tex].
2. Determine the range:
The output of the square root function [tex]\( \sqrt{x + 3} \)[/tex] is non-negative (i.e., [tex]\( \geq 0 \)[/tex]) for [tex]\( x \geq -3 \)[/tex]. The smallest value of [tex]\( \sqrt{x + 3} \)[/tex] occurs when [tex]\( x = -3 \)[/tex] and is zero. Subtracting 2 shifts the entire function down by 2 units. Thus, [tex]\( \sqrt{x+3} - 2 \)[/tex] starts at [tex]\(-2\)[/tex] when [tex]\( x = -3 \)[/tex] and increases without bound as [tex]\( x \)[/tex] increases. Therefore, the range of [tex]\( f(x) \)[/tex] is [tex]\( y \in [-2, \infty) \)[/tex].
3. Find the x-intercept:
To find the x-intercept, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \sqrt{x + 3} - 2 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ \sqrt{x + 3} = 2 \quad \Rightarrow \quad x + 3 = 4 \quad \Rightarrow \quad x = 1 \][/tex]
Thus, the x-intercept occurs at [tex]\( (1, 0) \)[/tex].
4. Find the y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{0 + 3} - 2 = \sqrt{3} - 2 \][/tex]
Since [tex]\( \sqrt{3} \approx 1.732 \)[/tex], we have:
[tex]\[ f(0) \approx 1.732 - 2 = -0.268 \][/tex]
Thus, the y-intercept is approximately [tex]\( (0, -0.268) \)[/tex].
5. Plot additional points:
Calculate a few more points within the domain to understand the function's behavior:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \sqrt{0} - 2 = -2 \][/tex]
- When [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \sqrt{1} - 2 = 1 - 2 = -1 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \sqrt{4} - 2 = 2 - 2 = 0 \][/tex]
- When [tex]\( x = 7 \)[/tex]:
[tex]\[ f(7) = \sqrt{10} - 2 \approx 3.162 - 2 = 1.162 \][/tex]
6. Sketch the graph:
Plot the points [tex]\((-3, -2), (1, 0), (0, -0.268), (-2, -1), (7, 1.162)\)[/tex] and draw a curve passing through these points keeping in mind the domain [tex]\( x \geq -3 \)[/tex]. The graph starts at the point [tex]\((-3, -2)\)[/tex] and rises to the right.
The graph of [tex]\( f(x) = \sqrt{x + 3} - 2 \)[/tex] is thus a transformed square root function that has been shifted 3 units to the left (because of [tex]\( x + 3 \)[/tex] under the square root) and 2 units down (because of the [tex]\(-2\)[/tex]). It opens to the right from the starting point [tex]\((-3, -2)\)[/tex].
So, the correct graph is the one that starts at [tex]\((-3, -2)\)[/tex], passes through [tex]\( (1, 0) \)[/tex], and continues to rise slowly as [tex]\( x \)[/tex] increases, following the standard shape of a square root function adjusted by the transformations described above.
