Lesson 4-5 Warm-Up

“Direct Variation” (4-5) What is a “direct variation”? How can you tell if an equation is a direct variation? Direct Variation (sometimes called a direct proportion): a linear (forms a line on a graph) function n the form of y = kx, where k ≠ 0 and coefficient k is called the “constant of the variation” (a number that never changes). This means that the y varies, or “changes”, directly, or proportionally, with changes in x. Note: Since y = 0 when x = 0, all direct variations pass through the origin (0, 0) Examples: y = ¾ x y = -½ x You can tell an equation is a direct variation if it is written in the form y = kx after you solve for y. Examples:

–3y = 1 – 2x Subtract 2x from each side. Direct Variation LESSON 4-5 Additional Examples Is each equation a direct variation? If it is, find the constant of variation. a. 2x – 3y = 1 –3y = 1 – 2x Subtract 2x from each side. y = – + x Divide each side by –3. 1 3 2 The equation does not have the form y = kx. It is not a direct variation. b. 2x – 3y = 0 –3y = –2x Subtract 2x from each side. y = x Divide each side by –3. 2 3 The equation has the form y = kx, so the equation is a direct variation. 2 3 The constant of variation is .

“Direct Variation” (4-5) How do you write the equation for direct variation? Tip: To write the equation for a direct variation, you need to determine what k is. You can find k using a point other than the origin (0,0) that lies on the graph of the direct variation and substituing the x and y values in y = kx. Once you find the value of k, simply replace it (y = __x). Example: Write a direct variation equation that includes the point (4, -3). An equation for the direct variation that includes (4, -3) is y = -¾x

y = kx Use the general form of a direct variation. LESSON 4-5 Additional Examples Write an equation for the direct variation that includes the point (–3, 2). y = kx Use the general form of a direct variation. 2 = k(–3) Substitute –3 for x and 2 for y. – = k Divide each side by –3 to solve for k. 2 3 y = – x Write an equation. Substitute – for k in y = kx. 2 3 The equation of the direct variation is y =– x . 2 3

Words: The weight varies directly with the mass. When x = 6, y = 59. Direct Variation LESSON 4-5 Additional Examples The weight an object exerts on a scale varies directly with the mass of the object. If a bowling ball has a mass of 6 kg, the scale reads 59. Write an equation for the relationship between weight and mass. Words: The weight varies directly with the mass. When x = 6, y = 59. Define: Let = the mass of an object. Let = the weight of an object. x y

Equation: = k Use the general form of a direct variation. x y LESSON 4-5 Additional Examples (continued) Equation: = k Use the general form of a direct variation. x y 59 = k(6) Solve for k. Substitute 6 for x and 59 for y. = k Divide each side by 6 to solve for k. 59 6 y = x Write an equation. Substitute for k in y = kx. 59 6 The equation y = x relates the weight of an object to its mass. 59 6

“Direct Variation” (4-5) How do you tell whether each “data pair” (x and y) in a table is a direct variation? Tip: You can write y = kx as k = y / x if you divide both sides by x. If each data pair (x  y) equals k (in other words, the ratio of y to x is the same for each x and y pair), then the table represents a direct variation. Example: Is the following table a direct variation? No, the ratio of (in other words, the “k”) is not the same for all of the x and y data pairs.

Yes, the constant of variation is –0.5. The equation is y = –0.5x. Direct Variation LESSON 4-5 Additional Examples For the data in each table, use the ratio to tell whether y varies directly with x. If it does, write an equation for the direct variation. y x 1 –2 = –0.5 –1 2 4 y x x y –2 1 2 –1 4 –2 2 –1 = –2 1 –4 = 2 y x x y –1 2 1 2 2 –4 a. b. Yes, the constant of variation is –0.5. The equation is y = –0.5x. No, the ratio is not the same for each pair of data. y x

Define: Let n = the force you need to lift 210 lb. Direct Variation LESSON 4-5 Additional Examples Suppose a windlass requires 0.75 lb of force to lift an object that weighs 48 lb. How much force would you need to lift 210 lb? Words: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb. Define:  Let n = the force you need to lift 210 lb. Equation: = Use a proportion. force1 weight1 force2 weight2 Substitute 0.75 for force1, 48 for weight1, and 210 for weight2. 0.75 48 n 210 = 0.75(210) = 48n Use cross products. Solve for n. n 3.3 You need about 3.3 lb of force to lift 210 lb.

Direct Variation LESSON 4-5 Lesson Quiz 1. Is each equation a direct variation? If it is, find the constant of variation. a. x + 5y = 10 b. 3y + 8x = 0 no yes; – 8 3 2. Write an equation of the direct variation that includes the point (–5, –4). y = x 4 5 3. For each table, tell whether y varies directly with x. If it does, write an equation for the direct variation. x y –1 3 0 0 2 –6 3 –9 a. x y –1 –2 0 0 1 2 3 –6 b. yes; y = – 3x no