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Solve for a.
2. –5a = –16
1. 10 = 2a
ANSWER
5
ANSWER
16
5
Solve for a.
3.
Write an equation of the line that passes through the
points (0, 0) and (4, 8).
y = 2x
Main
Concept
EXAMPLE 1
Write and graph a direct variation equation
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.
y = ax
Write direct variation equation.
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)
y = ax
-9 = a(3)
-3 = a
ANSWER
y = – 3x.
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
EXAMPLE 2
Write and apply a model for direct variation
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.
EXAMPLE 2
Write and apply a model for direct variation
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.
d = at
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.
Correlation: +1
Correlation: -1
Correlation: 0
Correlation: +1/2
Correlation: -1/2
EXAMPLE 1
Describe correlation
Describe the correlation shown by each scatter plot.
Positive
or
Negative
Positive Correlation
Negative Correlation
The 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.
EXAMPLE 2
Estimate correlation coefficients
Tell whether the correlation coefficient for the data is
closest 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.)
EXAMPLE 2
Estimate correlation coefficients
b.
SOLUTION
b. The scatter plot shows approximately no correlation.
So, the best estimate given is r = 0.
(The actual value is r
– 0.02.)
EXAMPLE 2
Estimate correlation coefficients
c.
SOLUTION
c.
The scatter plot shows a strong positive correlation.
So, the best estimate given is r = 1.
(The actual value is r 0.98.)
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
Alternative-fueled Vehicles
The table shows the number y
(in thousands) of alternative-fueled
vehicles in use in the United States x years
after 1997. Approximate the best-fitting
line for the data.
This will be very important for
our project next week!
x
0
1
2
3
4
5
6
7
y
280
295
322
395
425
471
511
548
EXAMPLE 3
Approximate a best-fitting line
SOLUTION
STEP 1
Draw a scatter plot of the data.
STEP 2
Sketch the line that appears to
best fit the data. One possibility
is shown.
EXAMPLE 3
Approximate a best-fitting line
STEP 3
Choose two points that appear to lie on the line. For the
line shown, you might choose (1,300), which is not an
original data point, and (7,548), which is an original data
point.
STEP 4
Write an equation of the line. First find the slope using
the points (1,300) and (7,548).
248
548 – 300
m=
=
6
7–1
41.3
EXAMPLE 3
Approximate a best-fitting line
Use point-slope form to write the equation. Choose
(x1, y1) = (1,300).
y – y1 = m(x – x1)
y – 300 = 41.3(x – 1)
y
41.3x + 259
Point-slope form
Substitute for m, x1, and y1.
Simplify.
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 to
predict the number of alternative-fueled vehicles in use
in the United States in 2010.
SOLUTION
Because 2010 is 13 years after 1997, substitute 13 for x
in the equation from Example 3.
y = 41.3x + 259 = 41.3(13) + 259
796
EXAMPLE 4
Use a line of fit to make a prediction
ANSWER
You can predict that there will be about 796,000
alternative-fueled vehicles in use in the United States in
2010.
Classwork:
Worksheet 2-5 (1-21 odd)
Worksheet 2-6 (1-10 all)