We are asked to verify if the given system of inequalities will always have at least one solution. In the exercise we claim that it wouldn't, but our friend claims that it would. Let's consider the given system of inequalities.
{y<(x+a)2y<(x+b)2
To graph the , we will follow three steps.
- Graph the related .
- Test a point not on the .
- Shade accordingly. If the point satisfies the inequality, we shade the region that contains the point. If not, we shade the opposite region.
Let's draw the graphs of the related functions, which are y=(x+a)2 and y=(x+b)2.
Next, let's determine which region to shade by testing a point. We will use points below the vertices. For instance, for the inequality
y<(x+a)2 we will use
(-a,-1) as our test point. Let's see if it satisfies the given inequality.
y<(x+a)2
-1<?(-a+a)2
-1<?(0)2
-1<0 ✓
Since
(-a,-1) produced a true statement, we will shade the region that contains the point. Also, note that the inequality is strict. Therefore, the parabola will be dashed.
Next, we will draw y<(x+b)2. Let's determine which region to shade by testing a point. We will use, similarly as before, the point (-b,-1) as our test point.
Inequality
|
Test point
|
Statement
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Is it shaded?
|
y<(x+b)2
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(-b,-1)
|
-1<0 ✓
|
Yes
|
Let's graph it!
The solution set is the overlapping region.
Since the parabolas open upwards, we will always find a non-empty region which is the solution of the given system of inequalities. Therefore, our friend is right.