Solution

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If we change the linear function so its graph doesn t intersect the parabola, then the resulting two-by-two system will have no real solutions Imagine moving the straight line downward in Fig 28-1 As we do this, the solution points get closer together, eventually merging At the moment the two points become one, the system has a single real solution with multiplicity 2 The linear function is expressed in slope-intercept form, so we can see that the line has a slope of 2 and a y-intercept of 1 Each division on the y axis is 5 units, so we can see that the quadratic function has an absolute minimum of approximately 5 Suppose we change the linear function so the y-intercept is 10 That will put the line completely below the parabola, and will give us the two-by-two system y = 2x 10 and y = x2 + x 5

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466 More Two-by-Two Graphs

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(3,7)

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( 2, 3)

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Figure 28-1 Graphs of y = 2x + 1 and y = x 2 + x 5 The first

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function is graphed as a solid line; the second function is graphed as a dashed curve Real-number solutions appear as points where the line and the curve intersect On the x axis, each increment is 1 unit On the y axis, each increment is 5 units

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If we solve this system, we ll get two results, but they ll both be non-real complex numbers For extra credit, you can solve the revised system and see for yourself

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Does the slope of the line in Fig 28-1 seem smaller than 2 In a way, it is! In the algebraic sense the slope is 2, but in the geometric sense it s only 2/5 The increments on the vertical axis are five times as large as the ones on the horizontal axis That distorts the slopes and contours of the graphs, expanding everything horizontally (or compressing everything vertically) by a factor of 5 If we had used a true Cartesian plane, the line would have the steepness we should expect for a slope of 2 when drawn The parabola would also look different; it would seem five times sharper

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Two Quadratics

In the second example in Chap 27, we solved this system of quadratic equations in two variables: 4x 2 + 6x + 2y + 8 = 0 and 3x 2 + y + 5x 11 = 0

Two Quadratics

Again, we let x be the independent variable, and then we manipulated the equations to obtain y as a function of x in both cases That gave us y = 2x 2 3x 4 and y = 3x 2 5x + 11

First, we tabulate some points Table 28-2 shows some values of x, along with the results of plugging those values into the above functions and churning out the arithmetic We can start building the table by entering the two solutions, which we determined in Chap 27 They are

(x, y) = ( 5, 39) and (x, y) = (3, 31) These solutions are written as bold numerals The other values are chosen to produce graph points in the vicinity of the solutions We can choose a couple of x-values less than 5, two more between 5 and 3, and two more larger than 3 Then we can calculate the values of the functions, and write them in the middle and right-hand columns of the table

Table 28-2 Selected values for graphing the functions y = 2x 2 3x 4 and y = 3x 2 5x + 11 Bold entries indicate real solutions

x 10 7 5 2 0 3 6 9 2x 2 3x 4 174 81 39 6 4 31 94 193 3x 2 5x + 11 239 101 39 9 11 31 127 277