1. {(-2, 3), (-1, 2), (0, 2), (-2, 9)}
a. Is it a function? Explain.
b. What is the domain?
c. What is the range?
2. Bowling costs $4.50 per game.
Shoe rental costs $3.75. Write a
function that models the total cost
of games and shoe rental. Evaluate
the function for 10 games.
1. {(-2, 3), (-1, 2), (0, 2), (-2, 9)}
a. Is not a function. When x =-2,
there are TWO y values.
b. Domain: {-2, -1, 0}
c. Range: {2, 3, 9}
2. 𝑓 𝑥 = 4.50𝑥 + 3.75
For 10 games
𝑓 10 = 4.50(10) + 3.75
= $48.75
Tues
9/22
Learning Objective:
To write & interpret
direction variation eq’ns
Lesson Hw: Pg. 71 #7, 9, 19 – 25
2–2
odd, 39 – 45 odd, CC*51,
52, 71
Algebra II
 To
write & interpret direct
variation equations
 Direct
Variation – If input (x)
increase, then output (y)
increase. If input (x) decrease,
then output (y) decreases.
y = kx
of variation k – the ratio
of ALL output-input pairs equal k
𝑦
k=
𝑥
 Constant
1. Determine whether y varies directly
with x.
k=
𝑦
𝑥
2
1
6
3
8
4
= = = =2
Yes, y varies directly with x.
The constant of variation
k is 2.
y = kx
y = 2x
2. Determine whether y varies
directly with x.
k=
𝑦
𝑥
4
1
8
2
= = ≠
11
3
No, y does not vary
directly with x.
𝑦
𝑥
is not constant
𝑦
𝑥
In a direct variation, is the same
for all pairs of data, so you can
solve a direct variation by a
proportion
𝑦1 𝑦2
=
𝑥1 𝑥2
3. Suppose y varies directly with x,
and y = 9, when x = -15. What is y
when x = 21.
𝑦1
𝑥1
=
𝑦2
𝑥2
9

−15
=
𝑦
21
-15y = 9(21)
-15y = 189
-15
-15
y =-12.6
4. Suppose y varies directly with x,
and y = 15, when x = 3. What is x
when y = 60.
𝑦1
𝑥1
=
𝑦2
𝑥2
15

3
=
60
𝑥
15x = 3(60)
15x = 180
15
15
x =12
5. Write a direct variation equation
that passes through (3, -15)
y = kx
-15 = k(3)
3
3
-5 = k
y = -5x
6. Write a direct variation equation
that passes through (-7, -5)
y = kx
-5 = k(-7)
-7
-7
5
7
=k
y=
5
𝑥
7
7. A salesperson’s commission varies
directly with sales. For $1000 in sales,
the commission is $85. What is the
commission for $2300 in sales?
y= commission
x = sales
𝑦1
𝑦2
85
𝑦
=

=
𝑥1
𝑥2
1000
2300
1000y = 85(2300)
1000y = 195500
1000
1000
y = $195.50
8. The number of calories varies
directly with the mass of cheese. If 50
grams of cheese contains 200
calories, how many calories are in 70
grams of cheese?
y= calories
x = cheese
𝑦1
𝑦2
200
𝑦
=

=
𝑥1
𝑥2
50
70
50y = 200(70)
50y = 14000
50
50
y = 280 calories
Algebra II
1. (-3, 7) & (-2, 4)
𝒚𝟐 − 𝒚𝟏
𝟒−𝟕
−𝟑
𝑚=
=
=
= −𝟑
𝒙𝟐 − 𝒙𝟏 −𝟐 − (−𝟑)
𝟏
Find parallel slope
-3
Find perpendicular slope
1
3
2. (3, 1) & (-4, 1)
𝒚𝟐 − 𝒚𝟏
𝟏−𝟏
𝟎
𝑚=
=
=
=𝟎
𝒙𝟐 − 𝒙𝟏 −𝟒 − 𝟑 −𝟕
horizontal line!
At y = 1
3. (7, 3) & (7, 1)
𝒚𝟐 − 𝒚𝟏 𝟏 − 𝟑 −𝟐
𝑚=
=
=
𝒙𝟐 − 𝒙𝟏 𝟕 − 𝟕
𝟎
= 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅
CANNOT
DIVIDE BY
vertical line!
at x = 7 ZERO!!
When slope is undefined, the graph
is a vertical line.
Parallel line would also be vertical
with a slope that is undefined.
Perpendicular line would be
horizontal with a slope that is 0.
When slope is 0, the graph is a
horizontal line.
Parallel line would also be
horizontal with a slope of 0.
Perpendicular line would be vertical
with a slope that is undefined.
 Explain
why you cannot answer
the following question.
If y = 0 when x = 0, what is x when
y = 13?