Answer :
To solve the equation
[tex]\[ 2|x-2| - 5 = \sqrt{x+3} - 1 \][/tex]
graphically, we'll first consider the graphs of the two sides of the equation independently:
1. [tex]\( y_1 = 2|x-2| - 5 \)[/tex]
2. [tex]\( y_2 = \sqrt{x+3} - 1 \)[/tex]
We'll look at the points of intersection of these two graphs to determine the values of [tex]\( x \)[/tex] that satisfy the equation. Let's sketch or plot these graphs:
Step 1: Graph [tex]\( y_1 = 2|x-2| - 5 \)[/tex]
- The absolute value function [tex]\( |x-2| \)[/tex] translates the graph of [tex]\( |x| \)[/tex] horizontally to the right by 2 units.
- The function [tex]\( 2|x-2| \)[/tex] then stretches this graph vertically by a factor of 2.
- Finally, subtracting 5 shifts the graph downward by 5 units.
The resulting graph forms a "V" shape, with the vertex at [tex]\( (2, -5) \)[/tex].
Step 2: Graph [tex]\( y_2 = \sqrt{x+3} - 1\)[/tex]
- The square root function [tex]\( \sqrt{x+3} \)[/tex] translates the graph of [tex]\( \sqrt{x} \)[/tex] horizontally to the left by 3 units.
- The function [tex]\( \sqrt{x+3} - 1 \)[/tex] then shifts this graph downward by 1 unit.
Step 3: Find Points of Intersection
1. From visual inspection of the graphs:
- The [tex]\( y_1 \)[/tex] function converges from two directions due to the absolute value term, creating a "V" shape with a vertex at [tex]\( (2, -5) \)[/tex].
- The [tex]\( y_2 \)[/tex] function starts at [tex]\(( -3, -1 ) \)[/tex] and increases slowly, following a square root curve shifted to the right by three units and down by one unit.
We need to find the approximate [tex]\( x \)[/tex]-values where these graphs intersect.
From the possible options, you can analyze:
- Graphically examining the [tex]\( y = 2|x-2| - 5 \)[/tex] and [tex]\( y = \sqrt{x+3} - 1 \)[/tex] curves, solutions points [tex]\( x \approx -0.50 \)[/tex] and [tex]\( x \approx 4.5 \)[/tex] commonly match the regions where these two graphs intersect, when analyzing these points in detail.
Thus, the correct answer is:
C. [tex]\( x \approx -0.50 \)[/tex] and [tex]\( x \approx 4.5 \)[/tex]
[tex]\[ 2|x-2| - 5 = \sqrt{x+3} - 1 \][/tex]
graphically, we'll first consider the graphs of the two sides of the equation independently:
1. [tex]\( y_1 = 2|x-2| - 5 \)[/tex]
2. [tex]\( y_2 = \sqrt{x+3} - 1 \)[/tex]
We'll look at the points of intersection of these two graphs to determine the values of [tex]\( x \)[/tex] that satisfy the equation. Let's sketch or plot these graphs:
Step 1: Graph [tex]\( y_1 = 2|x-2| - 5 \)[/tex]
- The absolute value function [tex]\( |x-2| \)[/tex] translates the graph of [tex]\( |x| \)[/tex] horizontally to the right by 2 units.
- The function [tex]\( 2|x-2| \)[/tex] then stretches this graph vertically by a factor of 2.
- Finally, subtracting 5 shifts the graph downward by 5 units.
The resulting graph forms a "V" shape, with the vertex at [tex]\( (2, -5) \)[/tex].
Step 2: Graph [tex]\( y_2 = \sqrt{x+3} - 1\)[/tex]
- The square root function [tex]\( \sqrt{x+3} \)[/tex] translates the graph of [tex]\( \sqrt{x} \)[/tex] horizontally to the left by 3 units.
- The function [tex]\( \sqrt{x+3} - 1 \)[/tex] then shifts this graph downward by 1 unit.
Step 3: Find Points of Intersection
1. From visual inspection of the graphs:
- The [tex]\( y_1 \)[/tex] function converges from two directions due to the absolute value term, creating a "V" shape with a vertex at [tex]\( (2, -5) \)[/tex].
- The [tex]\( y_2 \)[/tex] function starts at [tex]\(( -3, -1 ) \)[/tex] and increases slowly, following a square root curve shifted to the right by three units and down by one unit.
We need to find the approximate [tex]\( x \)[/tex]-values where these graphs intersect.
From the possible options, you can analyze:
- Graphically examining the [tex]\( y = 2|x-2| - 5 \)[/tex] and [tex]\( y = \sqrt{x+3} - 1 \)[/tex] curves, solutions points [tex]\( x \approx -0.50 \)[/tex] and [tex]\( x \approx 4.5 \)[/tex] commonly match the regions where these two graphs intersect, when analyzing these points in detail.
Thus, the correct answer is:
C. [tex]\( x \approx -0.50 \)[/tex] and [tex]\( x \approx 4.5 \)[/tex]
