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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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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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