Answer :
To determine which quadrants the parabola given by the equation [tex]\( x = 2y^2 + 3 \)[/tex] occupies, we can analyze the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### Step-by-Step Analysis:
1. Rewrite the Equation for Analysis:
The given equation is [tex]\( x = 2y^2 + 3 \)[/tex].
2. Determine the Sign of [tex]\( x \)[/tex]:
Since [tex]\( y^2 \)[/tex] is always non-negative (i.e., [tex]\( y^2 \geq 0 \)[/tex]), it follows that:
[tex]\[ x = 2y^2 + 3 \][/tex]
Here, the term [tex]\( 2y^2 \)[/tex] is always non-negative, and adding 3 to it ensures that [tex]\( x \)[/tex] is always positive:
[tex]\[ x > 0 \][/tex]
3. Analyze the Quadrants Based on [tex]\( y \)[/tex]:
- Quadrant I: [tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant II: [tex]\( x < 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant III: [tex]\( x < 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
- Quadrant IV: [tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
Since [tex]\( x \)[/tex] is always positive from the previous step, the parabola cannot occupy Quadrants II and III, where [tex]\( x \)[/tex] would need to be negative.
4. Determine Which Quadrants Fit:
- For Quadrant I ([tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]):
Since [tex]\( x \)[/tex] is definitively positive and [tex]\( y \)[/tex] can be any positive number, this quadrant is occupied.
- For Quadrant IV ([tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]):
Similarly, [tex]\( x \)[/tex] is positive and since [tex]\( y \)[/tex] can be any negative number, this quadrant is also occupied.
Therefore, the parabola [tex]\( x = 2y^2 + 3 \)[/tex] occupies the following quadrants:
- Quadrant I (1)
- Quadrant IV (IV)
Thus, the correct selection of quadrants is:
1, IV.
### Step-by-Step Analysis:
1. Rewrite the Equation for Analysis:
The given equation is [tex]\( x = 2y^2 + 3 \)[/tex].
2. Determine the Sign of [tex]\( x \)[/tex]:
Since [tex]\( y^2 \)[/tex] is always non-negative (i.e., [tex]\( y^2 \geq 0 \)[/tex]), it follows that:
[tex]\[ x = 2y^2 + 3 \][/tex]
Here, the term [tex]\( 2y^2 \)[/tex] is always non-negative, and adding 3 to it ensures that [tex]\( x \)[/tex] is always positive:
[tex]\[ x > 0 \][/tex]
3. Analyze the Quadrants Based on [tex]\( y \)[/tex]:
- Quadrant I: [tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant II: [tex]\( x < 0 \)[/tex] and [tex]\( y > 0 \)[/tex]
- Quadrant III: [tex]\( x < 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
- Quadrant IV: [tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]
Since [tex]\( x \)[/tex] is always positive from the previous step, the parabola cannot occupy Quadrants II and III, where [tex]\( x \)[/tex] would need to be negative.
4. Determine Which Quadrants Fit:
- For Quadrant I ([tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex]):
Since [tex]\( x \)[/tex] is definitively positive and [tex]\( y \)[/tex] can be any positive number, this quadrant is occupied.
- For Quadrant IV ([tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex]):
Similarly, [tex]\( x \)[/tex] is positive and since [tex]\( y \)[/tex] can be any negative number, this quadrant is also occupied.
Therefore, the parabola [tex]\( x = 2y^2 + 3 \)[/tex] occupies the following quadrants:
- Quadrant I (1)
- Quadrant IV (IV)
Thus, the correct selection of quadrants is:
1, IV.