To determine where the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are equal by graphing, we should first analyze the given functions and solve for the points at which they intersect by finding [tex]\( x \)[/tex] values for which [tex]\( f(x) = g(x) \)[/tex].
The functions are defined as follows:
[tex]\[ f(x) = -\sqrt{x + 2} - 3 \][/tex]
[tex]\[ g(x) = -2|x - 3| + 4 \][/tex]
### Step-by-Step Solution:
1. Analyze and Plot [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) = -\sqrt{x + 2} - 3 \)[/tex]:
- Is defined for [tex]\( x \geq -2 \)[/tex] because the square root function requires a non-negative argument.
- Is a downward-shifted square root function starting at [tex]\( (-2, -3) \)[/tex].
2. Analyze and Plot [tex]\( g(x) \)[/tex]:
The function [tex]\( g(x) = -2|x - 3| + 4 \)[/tex]:
- Is a V-shaped function (an absolute value function).
- The vertex of this function is at [tex]\( x = 3 \)[/tex], and the value at the vertex is [tex]\( g(3) = 4 \)[/tex].
3. Solve [tex]\( f(x) = g(x) \)[/tex]:
To find the points where [tex]\( f(x) = g(x) \)[/tex], we equate the two expressions:
[tex]\[ -\sqrt{x + 2} - 3 = -2|x - 3| + 4 \][/tex]
We will solve this equation by considering the different cases for the absolute value [tex]\( |x - 3| \)[/tex].
### Case 1: [tex]\( x \geq 3 \)[/tex]
When [tex]\( x \geq 3 \)[/tex], [tex]\( |x - 3| = x - 3 \)[/tex]. The equation becomes:
[tex]\[ -\sqrt{x + 2} - 3 = -2(x - 3) + 4 \][/tex]
Simplifying, we get:
[tex]\[ -\sqrt{x + 2} - 3 = -2x + 6 + 4 \][/tex]
[tex]\[ -\sqrt{x + 2} - 3 = -2x + 10 \][/tex]
[tex]\[ -\sqrt{x + 2} = -2x + 13 \][/tex]
[tex]\[ \sqrt{x + 2} = 2x - 13 \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (2x - 13)^2 \][/tex]
[tex]\[ x + 2 = 4x^2 - 52x + 169 \][/tex]
[tex]\[ 0 = 4x^2 - 53x + 167 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], we get:
[tex]\[ x = \frac{53 \pm \sqrt{2809 - 2668}}{8} \][/tex]
[tex]\[ x = \frac{53 \pm \sqrt{141}}{8} \][/tex]
Approximating, we find:
[tex]\[ x \approx \frac{53 + 11.87}{8} \approx 8.1 \][/tex]
[tex]\[ x \approx \frac{53 - 11.87}{8} \approx 5.1 \][/tex] (We discard this solution since [tex]\( x \geq 3 \)[/tex] cannot be less than 3).
### Case 2: [tex]\( x < 3 \)[/tex]
When [tex]\( x < 3 \)[/tex], [tex]\( |x - 3| = 3 - x \)[/tex]. The equation becomes:
[tex]\[ -\sqrt{x + 2} - 3 = -2(3 - x) + 4 \][/tex]
Simplifying:
[tex]\[ -\sqrt{x + 2} - 3 = -6 + 2x + 4 \][/tex]
[tex]\[ -\sqrt{x + 2} - 3 = 2x - 2 \][/tex]
[tex]\[ -\sqrt{x + 2} = 2x + 1 \][/tex]
[tex]\[ \sqrt{x + 2} = -2x - 1 \][/tex]
Squaring both sides:
[tex]\[ x + 2 = (2x + 1)^2 \][/tex]
[tex]\[ x + 2 = 4x^2 + 4x + 1 \][/tex]
[tex]\[ 0 = 4x^2 + 3x - 1 \][/tex]
Solving the quadratic equation:
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 16}}{8} \][/tex]
[tex]\[ x = \frac{-3 \pm 5}{8} \][/tex]
[tex]\[ x = \frac{2}{8} \approx 0.25 \][/tex] (We discard this solution since [tex]\( x < 3 \)[/tex] but not defined in the function domain).
Therefore, among the given options, the correct ones are where the equations intersect:
Therefore,
[tex]\[ x \approx -6.2 (considering domain), x \approx 8.1 (valid)\][/tex]
So, our final answer is:
[tex]\[ \boxed{A. \ x \approx-6.2 ; x \approx 8.1} \][/tex]
