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b=2 b

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(x0, y0) = ( 1, 1)

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a=2 x

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Each axis increment is 1 unit

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Figure B-11

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Illustration for the solution to Problem 8 in Chap 13 Each axis division represents 1 unit

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This equation is in the standard form for a parabola The coefficient of x 2 is positive, so we know that the parabola opens upward When we divide the negative of the coefficient for x by twice the coefficient for x 2, we get the x value of the vertex point, which we can call x0 We have x0 = ( 3)/(2 1) = 3/2 To find the y value of the vertex point (let s call it y0), we plug 3/2 into the function for x and grind out the arithmetic: y0 = (3/2)2 3 3/2 + 2 = 9/4 9/2 + 2 = 9/4 18/4 + 8/4 = (9 18 + 8)/4 = 1/4 Now we know that the coordinates of the vertex are (3/2, 1/4) Because the parabola opens upward, we know that this vertex is the absolute minimum point The right-hand side of the standard-form equation factors into a product of two cleancut binomials y = (x 1)(x 2)

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536 Worked-Out Solutions to Exercises: 11-19

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If we set y = 0 and then solve the resulting quadratic equation, we get two roots that tell us the x-intercepts of the curve The roots here can be read straight from the factors as x=1 or x=2 revealing that (1,0) and (2,0) lie on the parabola These two points are close together, and they re also close to the vertex It s difficult to draw a good image of the parabola based on these three points alone But we can find the y-intercept, which is farther out, to help us draw the curve When we plug in 0 for x, we get y = 02 3 0 + 2 = 0 0 + 2 = 2 indicating that (0,2) is on the parabola Now that we know the coordinates of four points that lie on the parabola, we can do some mental curve-fitting and sketch the graph as shown in Fig B-12

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(0, 2)

Each axis increment is 1/4 unit

(1, 0) x (2, 0) (3/2, 1/4)

Figure B-12

Illustration for the solution to Problem 9 in Chap 13 Each axis division represents 1/4 unit

13

10 The standard-form general equation for a hyperbola in the xy plane is (x x0)2 /a2 (y y0)2 /b2 = 1 where (x0,y0) are the coordinates of the center, a is the length of the horizontal semi-axis, and b is the length of the vertical semi-axis The specific equation of our hyperbola is (x 1)2 /4 (y + 2)2 /9 = 1 From this equation, we can see that the constants are x0 = 1 y0 = 2 a = 41/2 = 2 b = 91/2 = 3 This information tells us that the hyperbola s center is at (1, 2) The horizontal semiaxis is 2 units wide The vertical semi-axis is 3 units tall We can therefore sketch the graph as shown in Fig B-13

Each axis increment is 1 unit

(x0, y0) = (1, 2)

Figure B-13

Illustration for the solution to Problem 10 in Chap 13 Each axis division represents 1 unit

538 Worked-Out Solutions to Exercises: 11-19

To find the equations of the asymptotes, we must know the coordinates of a point on each of them, and we must also know their slopes The asymptotes intersect at the center of the hyperbola, which is at (x0,y0) = (1, 2) We know that this point lies on both asymptote lines Let s construct a rectangle D whose width is twice the length of the horizontal semi-axis, and whose height is twice the length of the vertical semi-axis The asymptotes pass through the corners of this rectangle, as shown in Fig B-13 When we go from the center to the upper right-hand corner of D, the rise over run (an informal term for slope) is b/a = 3/2 so m = 3/2 We can therefore write the equation of the up-ramping asymptote in point-slope form as y + 2 = (3/2)(x 1) When we travel from the center to the lower right-hand corner of D, the rise over run is b/a = 3/2 so m = 3/2 We can therefore write the equation of the down-ramping asymptote in point-slope form as y + 2 = ( 3/2)(x 1) If you insist on having the equations of these lines appear in the standard form for a linear equation in two-space, feel free to convert them to that form!