Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Exam 2 Review 1 2.1-2.5, 3.1-3.4 2.1:Coordinate Geometry • Memorize 2 formulas for two points (x1 , y1 ) and (x2 , y2 ): Midpoint formula: x1 + x2 y1 + y2 , 2 2 p (x2 − x1 )2 + (y2 − y1 )2 Distance formula: d= 1. Find the midpoint and distance between (−3, 4) and (6, −2). Solution: Midpoint: 3 2, 1 √ , Distance: 3 13 2. Find the midpoint and distance between Solution: Midpoint: 13 −9 4 , 2 7 2 , −5 and (3, −4). √ , Distance: 5 2 2.2: Linear Equations and Inequalities • Be able to find x and y intercepts and be able to describe in words what x and y intercepts are. 3. Graph 6x + 8y ≤ −12. Solution: x intercept at (−2, 0) and y intercept at (0, −3/2). Connect the intercepts using a solid line. Shade the half-plane that does not include the origin (0, 0). 4. Graph 4x − 3y < 16. . Connect the intercepts using a dotted line. Solution: x intercept at (4, 0) and y intercept at 0, −16 3 Shade the half-plane that include the origin (0, 0). 2.3: Determining the Equation of a Line 2 • Memorize characteristics of 3 types of equations of lines: – Point-Slope Form: (y − y1 ) = m(x − x1 ) ◦ m is the slope ◦ x and y are the most general x and y ◦ (x1 , y1 ) is any specific point on the line – Slope-Intercept Form: y = mx + b ◦ m is the slope ◦ x and y are the most general x and y ◦ b is the y-intercept – Standard Form: Ax + By = C ◦ A, B, and C are all integers, A ≥ 0 ◦ x and y are the most general x and y • Memorize Property 2.1 – Two lines with slopes m1 and m2 are: ◦ parallel if m1 = m2 ◦ perpendicular if m1 m2 = −1 5. Find the equation of the line determined by the points (3, 3) and (5, −7). Put your answer in slopeintercept form. Solution: y = −5x + 18 6. Find the equation of the line which has y intercept −2 and slope 3/4. Put your answer in standard form. Solution: 3x − 4y = 8 7. Find the equation of the line parallel to 4x − 5y = 15 that goes through the point (6, 13). Put your answer in point-slope form. Solution: (y − 13) = 45 (x − 6) 8. Find the equation of the line perpendicular to y = − 85 x − 1 through the point (2, 1). Put your answer in standard form. Solution: 8x − 5y = 11 2.4: Graphing Techniques 3 • Memorize rules for determining symmetry – A graph has y-axis symmetry if replacing x with -x results in an equivalent equation – A graph has x-axis symmetry if replacing y with -y results in an equivalent equation – A graph has origin symmetry if replacing both x with -x and y with -y results in an equivalent equation • Be familiar with common restrictions: – Avoid √ anything and negative 0 • Look at resulting values in common graphs: √ – ◦ positive ≥ 0 – ◦ (anything)2 ≥ 0 – ◦ |anything| ≥ 0 9. Graph y = −2|x − 3| + 1. Solution: • No symmetry • x intercepts at ( 52 , 0) and ( 27 , 0), y intercept at (0, −5) • No restrictions on x. y ≤ 1 • The graph of y = |x| is shifted reflected across the x-axis and becomes twice as narrow. Then it is shifted 1 unit up and 3 units right. 10. Graph y = √ x2 + x − 2. Solution: • No symmetry • x intercepts at (−2, 0) and (1, 0), no y intercept • x ≤ −2 or x ≥ 1 and y ≥ 0 • You have to plot points since there isn’t really a basic graph we’re working with here. 11. Graph y = x+1 . x2 + x − 6 Solution: • No symmetry • x intercept at (−1, 0), y-intercept at 0, − 16 • Restrictions are x 6= 2 and x 6= −3. • You have to plot points since there isn’t really a basic graph we’re working with here. 4 2.5: Circles • Memorize the standard form for the equation of a circle with center at (h, k) and radius r. (x − h)2 + (y − k)2 = r2 • No ellipses or hyperbolas will be on the exam • Complete the square with two variables 12. Write the equation for this circle in standard form. Give the center and radius then graph the circle. 2x2 − 16x + 2y 2 + 20y + 58 = 0 √ Solution: Center at (4, −5) and radius 2 3 13. Write the equation for this circle in standard form. Give the center and radius then graph the circle. 1 2 1 x − 6x + y 2 + 2y + 36 = 0 4 4 Solution: Center at (12, −4) and radius 4 14. Write the equation for this circle in standard form. Give the center and radius then graph the circle. x2 − x + y 2 + 3y − Solution: Center at 1 3 2, −2 √ and radius 2 2 11 =0 2 3.1: Functions 5 • Domain is the set of all possible inputs x • Range is the set of all possible outputs f (x) • A function is even if it has y-axis symmetry. That is, a function is even if f (−x) = f (x) • A function is odd if it has origin symmetry. That is, a function is odd if f (−x) = −f (x). 15. Give the domain and range of the following functions. Is each function odd, even, or neither? √ ◦ f (x) = x − 2 Solution: Domain: {x|x ≥ 2}, Range: {f (x)|f (x) ≥ 0}, neither odd nor even ◦ g(x) = (x − 1)3 Solution: Domain: All real numbers, Range: All real numbers, neither odd nor even ◦ h(x) = √ x2 + 1 − 4 Solution: Domain: All real numbers, Range: {f (x)|f (x) ≥ −3}, even 3.2: Linear Functions • Be comfortable graphing a line in slope-intercept form • Be able to explain what f (x) means and what f (2) means 16. Graph f (x) = − 14 x + 7. What is f (4)? Solution: The y intercept is at (0, 7) and the x intercept is at (28, 0). f (4) = 6 17. Stephen makes $500 each month plus $50 for every cell phone he sells. Write a linear function that describes his situation. How much money does he make if he sells 4 cell phones? What is the domain? Solution: f (x) = 50x + 500. He will make $700 if he sells 4 cell phones. The domain is non-negative integers (0, 1, 2, 3, ...) since he can’t sell 3.5 or −2 cell phones. 3.3: Quadratic Functions 6 • Be able to explain what each part of vertex form y = a(x − h)2 + k means ◦ y = x2 + k is a vertical shift ◦ y = ax2 is a stretch, scrunch, or flip depending on the value of a ◦ y = (x − h)2 is a horizontal shift • Be able to graph a parabola in vertex form • Complete the square to write f (x) = ax2 + bx + c in vertex form 18. Graph f (x) = −(x − 5)2 + 3 Solution: The basic graph of y = x2 is flipped across the x axis so it now opens downward, moved 5 units right and 3 units up. 19. The basic graph of f (x) = x2 is scrunched (made narrower) by a factor of 2, is moved 6 units down, and 3 units left. Write the new quadratic function this new graph illustrates. Solution: f (x) = 2(x + 3)2 − 6 20. Put f (x) = 3x2 + 6x − 1 in vertex form then graph the function. Solution: f (x) = 3(x + 1)2 − 4. The parabola is made narrower by a factor of 3, moved 4 units down and 1 unit right. 3.4: More Quadratic Functions 7 • Be able to graph a parabola using the techniques from this chapter • Memorize the formula for the vertex of the parabola f (x) = ax2 + bx + c b 4ac − b2 − , 2a 4a • Be able to find the minimum or maximum value of a function based on a story problem 21. What is the vertex of f (x) = −2x2 − 4x + 2? Does this vertex represent a minimum or a maximum? Solution: The vertex is at a<0 −4 − 2(−2) , 4(−2)(2)−(−4) 4(−2) 2 = (−1, 4) and it represents a maximum since 22. You have 1200 feet of fencing available and you want to make a rectangular pen which is divided into 5 smaller pens of equal size. The fencing used to divide the larger area into smaller pens must be parallel to 2 sides of the rectangle. Write a quadratic equation that illustrates this situation. What are the dimensions of the rectangular pen that will maximize the area? Solution: The total length of fence is 1200 feet. This will build four sides of a rectangle plus four lines subdividing the rectangle, for a total of two sides of one length and six sides of a second length. This gives 6a + 2b = 1200. Solving for b, we get b = 21 (1200 − 6a) = 600 − 3a. So the two sides are a and 600−3a. The area of the pen is given by f (a) = a(600−3a) = −3a2 +600a. The vertex is at (100, 30000) and so the value of a that maximizes the area is a = 100. Then b = 600 − 3a = 600 − 300 = 300, so the dimensions of the pen that miximize the area are 100 feet by 300 feet.