Lesson 5-5Direct Variation

Lesson 5-5Direct Variation

Algebra I, Ms. Turk

Definition: direct variation

A direct variation is a function in the form y = kx where k does not equal 0.

Definition: constant of variation

The constant of variation is k, the coefficient of x.

An equation is a direct variation if:

its graph is a line that passes through zero, or

the equation can be written in the form y = kx.

Is an Equation a Direct Variation?If it is, find the constant of variation.

5x + 2y = 0 Solve for y.1. Subtract 5x.2. Divide by 2.

Yes, it’s a direct variation.Constant of variable, k, is

y 5

2x

5

2.

Is an Equation a Direct Variation?If it is, find the constant of variation.

5x + 2y = 9 Solve for y.1. Subtract 5x.2. Divide by 2.

No, it’s not a direct variation.It’s not in the form y = kx.

y 9 5x2

y 4.5 5

2x

Is an Equation a Direct Variation?If it is, find the constant of variation.

7y = 2x

Yes, it is a direct variation.The constant of variation, k, is

y 2

7x

2

7.

Writing an Equation Given a PointWrite an equation of the direct variation that

includes the point (4, -3).

y = kx Start with the function form of the direct variation.

-3 = k(4) Substitute 4 for x and -3 for y.

Divide by 4 to solve for k.

Substitute the value of k into the original formula.

k 3

4

y 3

4x

Writing an Equation Given a PointWrite an equation of the direct variation that

includes the point (-3, -6).

y = kx Start with the function form of the direct variation.

-6 = k(-3) Substitute -3 for x and -6 for y.

Divide by -3 to solve for k.

Substitute the value of k into the original formula.

k 2

y 2x

Real-World Problem Solving

Your distance from lightning varies directly with the time it takes you to hear thunder. If you hear thunder 10 seconds after you see the lightning,

you are about 2 miles from the lightning. Write an equation for the

relationship between time and distance.

Real-World Problem SolvingRelate: The distance varies directly with the time. When x = 10, y = 2.

Define: Let x = number of seconds between seeing lightning and hearing thunder.Let y = distance in miles from lightning.

Real-World Problem Solving

y = kx Use general form of direct variation.

2 = k(10) Substitute 2 for y and 10 for x.

Solve for k.

Write an equation using the value for k.

k 1

5

y 1

5x

Real-World Problem SolvingA recipe for a dozen corn muffins

calls for 1 cup of flower. The number of muffins varies directly with the amount of flour you use. Write a

direct variation for the relationship between the number of cups of flour

and the number of muffins.

Real-World Problem Solving

y = kx12 = k(1)

k = 12y = 12x

Real-World Problem Solving

The force you must apply to lift an object varies directly with the object’s

weight. You would need to apply 0.625 lb of force to a windlass to lift a 28-lb weight. How much force would you

need to lift 100 lb?

Relate: A force of 0.625 lifts 28 lb. What lifts 100 lb?

Use a proportion.

Cross multiply. 28n = 62.5Solve for n. n ≈ 2.2

You need about 2.2 lb of force to lift 100 lb.

0.625

28n

100

Real-World Problem Solving

Suppose a second windlass requires 0.5 lb of force to lift an object that weighs 32 lb. How much force would you

need to lift 160 lb?

Use a proportion.

Cross multiply. 32n = 80Solve for n. n = 2.5

You need 2.5 lb of force to lift 160 lb.

0.5

32n

160