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x2 4 1 0 1 4 x 2 4 1 0 1 4
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x 2 1 0 1 2
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Part Three 531
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(0,0)
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Figure 30-3 Illustration for Answer 28-2 The first function
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is graphed as a solid curve; the second function is graphed as a dashed curve The real-number solution appears as a point where the curves intersect On both axes, each increment is 1 unit
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the second function Figure 30-3 shows the curves and the solution point On both axes, each increment represents 1 unit
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Question 28-3
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Consider the system of equations we solved in Answer 27-3: y = x2 1 and y = x 2 + 1 How can we sketch an approximate graph of this system, showing the two real solutions
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Answer 28-3
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We can tabulate and plot several points in both functions including the real solutions, (1,0) and ( 1,0) Table 30-2 compares some values of x, some values of the first function, and some values of the second function Figure 30-4 shows the curves and the solution points On both axes, each increment represents 1 unit
532 Review Questions and Answers
Table 30-2
Selected values for graphing the functions y = x 2 1 and y = x 2 + 1 Bold entries indicate real solutions
x2 1 3 0 1 0 3 x 2 + 1 3 0 1 0 3
x 2 1 0 1 2 Question 28-4
When we compare the systems stated in Questions 28-2 and 28-3 and graphed in Figs 30-3 and 30-4, we can see that the curves have the same shapes in both situations But in the second case, the upward-opening parabola has been moved vertically down by 1 unit, while the downward-opening parabola has been moved vertically up by 1 unit This has caused the single intersection point (Fig 30-3) to break in two (Fig 30-4) What will happen if we move the upward-opening parabola, shown by the solid curve, further down, and move the downward-opening parabola, shown by the dashed curve, further up by the same distance How will the equations in the system change if we do this
Answer 28-4
If we move the parabolas this way, the intersection points will move farther from each other The negative x-value of one real solution will become more negative, and the positive x-value
( 1,0)
(1,0)
Figure 30-4 Illustration for Answer 28-3 The first function
is graphed as a solid curve; the second function is graphed as a dashed curve Real-number solutions appear as points where the curves intersect On both axes, each increment is 1 unit
Part Three 533
Solution
Solution
Figure 30-5 Illustration for Answer 28-4 The first function
is graphed as a solid curve; the second function is graphed as a dashed curve Real-number solutions appear as points where the curves intersect On both axes, each increment is 1 unit
of the other real solution will become more positive to the same extent The y-values of both solutions will remain at 0; the points will stay on the x axis Figure 30-5 shows an example On both axes, each increment represents 1 unit The stand-alone constants in the equations will change The negative constant in the first equation will become more negative, and the positive constant in the second equation will become more positive to the same extent
Question 28-5
Let s modify the system presented in Question 28-4 and graphed in Fig 30-5 Suppose that we move the upward-opening parabola even further straight down, but leave the downwardopening parabola in the same place What will happen to the solution points How will the equations in the system change
Answer 28-5
The intersection points will move even farther from each other The negative x-value of one real solution will become more negative, and the positive x-value of the other real solution will become more positive to the same extent The y-values of both solutions will become negative to an equal extent The solution points will move off the x axis into the third and fourth quadrants of the coordinate plane This assumes that we move the upward-opening parabola exactly in the negative-y direction Figure 30-6 shows an example On both axes, each increment represents 1 unit The negative constant in the first equation will become more negative, and the positive constant in the second equation will stay the same
Question 28-6
Consider the system of equations we solved in Answer 27-6: y = (x + 1)2
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