Solve for a.

1. 10 = 2a 2. –5a = –16

ANSWER 5ANSWER 16

5

3. Write an equation of the line that passes through the points (0, 0) and (4, 8).

y = 2x

Solve for a.

Main Concept

Write and graph a direct variation equation

EXAMPLE 1

Write and graph a direct variation equation that has (–4, 8) as a solution.

Use the given values of x and y to find the constant of variation.

Write direct variation equation.y = ax

8 = a(–4) Substitute 8 for y and – 4 for x.

–2 = a Solve for a.

Substituting –2 for a in y = ax gives the direct variation equation y = –2x.

GUIDED PRACTICE for Example 1

Write and graph a direct variation equation that has the given ordered pair as a solution.

1. (3, –9)

ANSWER

y = – 3x.

y = ax

-9 = a(3)-3 = a

GUIDED PRACTICE for Example 1

Write and graph a direct variation equation that has the given ordered pair as a solution.2. (–7, 4)

y = ax

4 = a(-7)-4/7 = a

y = -4/7x

Write and apply a model for direct variation

EXAMPLE 2

Meteorology

Hailstones form when strong updrafts support ice particles high in clouds, where water droplets freeze onto the particles. The diagram shows a hailstone at two different times during its formation.

Write and apply a model for direct variation

EXAMPLE 2

a. Write an equation that gives the hailstone’s diameter d (in inches) after t minutes if you assume the diameter varies directly with the time the hailstone takes to form.

b. Using your equation from part (a), predict the diameter of the hailstone after 20 minutes.

d = at

0.75 = a(12)

0.0625 = a An equation that relates t and d is d = 0.0625t.

After t = 20 minutes, the predicted diameter of the hailstone is d = 0.0625(20) = 1.25 inches.

d = at

                       

                       

                       

                       

                       

                       

                       

                       

                       

                       

                       

                       

                       

Correlation: +1Correlation: -1Correlation: 0Correlation: +1/2

Correlation: -1/2

Describe correlation

EXAMPLE 1

Describe the correlation shown by each scatter plot.

Positive Correlation Negative CorrelationThe first scatter plot shows a positive correlation, because as the number of cellular phone subscribers increased, the number of cellular service regions tended to increase .

The second scatter plot shows a negative correlation, because as the number of cellular phone subscribers increased, corded phone sales tended to decrease.

Positive

or

Negative

EXAMPLE 2 Estimate correlation coefficients

Tell whether the correlation coefficient for the data isclosest to – 1, – 0.5, 0, 0.5, or 1. a.

SOLUTION

a. The scatter plot shows a clear but fairly weak negative correlation. So, r is between 0 and – 1, but not too close to either one. The best estimate given is r = – 0.5. (The actual value is r –0.46.)

Estimate correlation coefficients

EXAMPLE 2

b. The scatter plot shows approximately no correlation. So, the best estimate given is r = 0. (The actual value is r – 0.02.)

SOLUTION

b.

Estimate correlation coefficients

EXAMPLE 2

c.

c. The scatter plot shows a strong positive correlation. So, the best estimate given is r = 1. (The actual value is r 0.98.)

SOLUTION

GUIDED PRACTICE for Examples 1 and 2

For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, – 0.5, 0, 0.5, or 1.

1. ANSWER

(a) positive correlation

(b) r = 0.5

GUIDED PRACTICE for Examples 1 and 2

For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, – 0.5, 0, 0.5, or 1.

2. ANSWER

(a) negative correlation

(b) r = –1

GUIDED PRACTICE for Examples 1 and 2

For each scatter plot, (a) tell whether the data have a positive correlation, a negative correlation, or approximately no correlation, and (b) tell whether the correlation coefficient is closest to –1, –0.5, 0, 0.5, or 1.

3. ANSWER

(a) no correlation

(b) r = 0

EXAMPLE 3 Approximate a best-fitting line

The table shows the number y(in thousands) of alternative-fueledvehicles in use in the United States x yearsafter 1997. Approximate the best-fittingline for the data.

x 0 1 2 3 4 5 6 7

y 280 295 322 395 425 471 511 548

Alternative-fueled Vehicles

This will be very important for our project next week!

Approximate a best-fitting line

EXAMPLE 3

SOLUTION

STEP 1

Draw a scatter plot of the data.

STEP 2

Sketch the line that appears tobest fit the data. One possibility is shown.

Approximate a best-fitting line

EXAMPLE 3

STEP 3

Choose two points that appear to lie on the line. For theline shown, you might choose (1,300), which is not anoriginal data point, and (7,548), which is an original datapoint.

STEP 4

Write an equation of the line. First find the slope usingthe points (1,300) and (7,548).

248 6

m = = 41.3548 – 300

7 – 1

Approximate a best-fitting line

EXAMPLE 3

Use point-slope form to write the equation. Choose(x1, y1) = (1,300).

y – y1 = m(x – x1) Point-slope form

y – 300 = 41.3(x – 1) Substitute for m, x1, and y1.

Simplify.y 41.3x + 259

ANSWER

An approximation of the best-fitting line is y = 41.3x + 259.

EXAMPLE 4 Use a line of fit to make a prediction

Use the equation of the line of fit from Example 3 topredict the number of alternative-fueled vehicles in usein the United States in 2010.

SOLUTION

Because 2010 is 13 years after 1997, substitute 13 for xin the equation from Example 3.

y = 41.3x + 259 = 41.3(13) + 259 796

Use a line of fit to make a prediction

EXAMPLE 4

ANSWER

You can predict that there will be about 796,000 alternative-fueled vehicles in use in the United States in 2010.

Classwork:Classwork: Worksheet 2-5 (1-21 odd)                    

Worksheet 2-6 (1-10 all)