MTH103 Section 065 Exam 2. x 2 + 6x + 7 = 2. x 2 + 6x + 5 = 0 (x + 1)(x + 5) = 0

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1 Absolute Value 1. (10 points) Find all solutions to the following equation: x 2 + 6x + 7 = 2 Solution: You first split this into two equations: x 2 + 6x + 7 = 2 and x 2 + 6x + 7 = 2, and solve each separately. x 2 + 6x + 7 = 2 x 2 + 6x + 5 = 0 (x + 1)(x + 5) = 0 We get from this the solutions x = 1 and x = 5. Next we solve the second equation: x 2 + 6x + 7 = 2 x 2 + 6x + 9 = 0 (x + 3) 2 = 0 From this we get the single solution x = 3. We have found three solutions, so the solution set is { 5, 3, 1}. 2. (10 points) Solve the following inequality and write the solution in interval notation. 6x 3 > 9 Solution: This statement is equivalent to saying 6x 3 < 9 or 6x 3 > 9. So we solve each of these inequalities: 6x 3 < 9 6x < 6 x < 1 The solution here is the interval (, 1). Now we solve the second inequality: 6x 3 > 9 6x > 12 x > 2 From this we get (2, ). So our end solution is described by x < 1 or x > 2, and is written in interval notation as the union (, 1) (2, ).

2 Equations of Circles and Lines 3. (12 points) Find the equations (in slope-intercept form) of the lines satisfying the following conditions: (a) The line has x-intercept of ( 6, 0) and y-intercept of (0, 3). 3 0 Solution: The slope is given by 0 ( 6) = 3 6 = 1 2. We already know the y-intercept is 3. So the equation in slope-intercept form is y = x (b) The line goes through the point (1, 1) and is parallel to the line 2x y = 3. Solution: The line y = 2x 3 has slope 2, so our line should also have slope 2. The point-slope formula says our line is given by y 1 = 2(x 1). If we multiply this out and move the 1 to the right side, we get the equation in slope-intercept form as y = 2x 1. (c) The line goes through the point (3, 4) and is perpendicular to the line y = 0. Solution: The line y = 0 is horizontal, so any line perpendicular to it is vertical. So our line needs to be the vertical line that goes through (3, 4). Since any vertical line has an equation that looks like x = c, it must be that our line is x = 3.

3 4. (10 points) The points (0, 0) and (6, 8) are the endpoints of the diameter of a circle. Determine the center and radius of the circle, and come up with an equation for the circle in standard form. Solution: The center is just the midpoint of the diameter (the segment joining (0, 0) and (6, 8)). This is given by ( , ) = (3, 4) (Center) 2 The radius is the distance from the center to any point on the circle. We know the center is (3, 4) now, and the point (0, 0) is on the circle. So the radius is the distance = 5. The equation (in standard form) is then (x 3) 2 + (y 4) 2 = 25 Polynomial and Rational Inequalities 5. (10 points) Solve the following polynomial inequality and write your solution in interval notation. (x + 5)(x 2)(x 3) > 0 Solution: The right side is already zero, and the left side is factored, so we can immediately see that our boundary points are x = 5, x = 2, and x = 3. So we have four intervals to test: Interval Test Point Evaluate (, 5) 6 ( 6 + 5)( 6 2)( 6 3) < 0 ( 5, 2) 0 (5)( 2)( 3) > 0 (2, 3) 2.5 (7.5)(0.5)( 0.5) < 0 (3, ) 4 (4 + 5)(4 2)(4 3) > 0 We see from this that our solution is ( 5, 2) (3, ).

4 6. (10 points) Solve the following rational inequality and write your solution in interval notation. 4x + 5 x 3 21 Solution: We first need to move the 21 to the other side and get everything into a single fraction before we can do anything: 4x + 5 x x + 5 x x (x 3) 0 x x x 3 Now we can see that the boundary points are the solutions to the equations 68 17x = 0 and x 3 = 0. These solutions are x = 4 and x = 3. So we have 3 intervals to test: 0 Interval Test Point Evaluate (, 3) < (3, 4) > 0 17 (4, ) 5 2 < 0 We are looking for when this is less than or equal to zero, so we see our solution consists of the intervals (, 3) and (4, ). The only remaining question is if we want the endpoints. We definitely don t want to include the endpoint 3, since x 3 is in the denominator of the original inequality. But since the inequality sign was, we want to include values where it equals zero, so we do include 4. So we get our solution to be (, 3) [4, )

5 7. (10 points) Find the average rate of change of the function y = x(x+4)(x 3) 8 from x = 5 to x = 4. Solution: We need to find y when x = 5 and when x = 4, and find the slope between these two points. When x = 5, we get y = 5( 1)( 8) 8 = 5. And when x = 4, we get y = 4(8)(1) 8 = 4. So our points are (x 1, y 1 ) = ( 5, 5) and (x 2, y 2 ) = (4, 4), and we need to take the slope of the line between them. This is given by y 2 y 1 = 4 ( 5) x 2 x 1 4 ( 5) = 9 9 = 1 8. (10 points) Compute the difference quotient, f(x+h) f(x) h, for the function f(x) = x Solution: First let s compute f(x + h): Next let s subtract the original function: Finally, we divide by h: f(x + h) = (x + h) = x 2 + 2hx + h f(x + h) f(x) = x 2 + 2hx + h (x 2 + 1) = 2hx + h 2 f(x + h) f(x) h = 2hx + h2 h = 2x + h

6 9. (6 points) True or False The slope of the line between the points (a, b) and (c, b) is zero. This is true. If you calculate the slope, you get b b c a = 0 c a = 0. The radius of the circle described by x 2 + (y 3) 2 = 4 is 2. This is true. This equation for a circle is in standard form, so the radius is the square root of the number on the right side of the equation. If a > b, then 2a > 2b. This is false. The correct statement would be that 2a < 2b. You have to flip the inequality sign if you multiply by a negative number. Multiple Choice The solution to the inequality x a < r is: (a r, a + r) A line parallel to x = 1: is vertical The solution set of the equation 2x 1 = 5 is: the empty set (none of these) The slope of the line perpendicular to y = 3 x is: 1

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