Pg164 3.5 Solving Problems Involving Quadratic Pg164
3.5
Problems involving quadratic functions can be solved with different strategies, such as
a table of values
graphing (with or without graphing technology)
an algebraic approach involving factoring
The strategy used depends on what you need to determine and whether or not an estimate or
a more accurate answer is required.
Note:
•
The value of in the quadratic equation
determines the direction in which the parabola
opens and whether there is a maximum or a
minimum. If &gt; 0, the parabola opens up and
there will be a minimum. If &lt; 0, the
parabola opens down and there will be a
maximum.
•
The quadratic function can be used to
determine the maximum or minimum value
and/or the zeros of the quadratic model.
These can then be used as needed to interpret
or solve the situation presented.
Example 1
A computer software company models the profit on its latest video game using the function
= −2 + 32 − 110, where is the number of games, in thousands, that the company
produces and is the profit, in millions of dollars. Determine the maximum profit.
= −2 + 32 − 110
= −2 − 16 + 55
0
−2 − 5 − 11
=
ℎ &divide; −2
−2
−2
0 = − 5 − 11
− 5 = 0 − 11 = 0
= 5 = 11
ℎ=
5 + 11 16
=
=8
2
2
8 = −28 + 328 − 110 = −128 + 256 − 110 = 18
The maximum profit is \$18, 000,000.
Example 2
Sally is standing on the top of a river slope and throws a ball. The height of the ball at a given
time is modeled by the function ℎ = −5 − 10 + 250, where ℎ is the height in meters
and is time in seconds. When will the ball be 10 m above the ground?
ℎ = −5 − 10 + 250
10 = −5 − 10 + 250
5 + 10 − 250 + 10 = 0
5 + 10 − 240 = 0
+ 2 − 48 = 0
− 6 + 8 = 0
−6=0
+8=0
=6
= −8 , ≥ 0
The ball will be 10 m above the ground at 6 s.
Homework
Pg169 #8~10
#8.
ℎ = −0.15 + 3
10 = −0.15 + 3
0.15 − 3 + 10 = 0
= 0.15,
= −3,
\$ = 10
=
− &plusmn; √ − 4\$
2
=
3 &plusmn; '−3 − 40.1510
20.15
=
3 &plusmn; √9 − 6 3 &plusmn; √3
=
0.3
0.3
) =
3 + √3
≅ 15.77
0.3
=
3 − √3
≅ 4.23
0.3
#9.
Method 1: complete the square
Method 2: factoring
= −2\$ + 14\$ − 20
= −2\$ + 14\$ − 20
= −2\$ − 7\$ + 3.5 − 3.5 − 20
0 = −2\$ + 14\$ − 20
= −2\$ − 7\$ + 3.5 − 2−3.5 − 20
0 = −2\$ − 7\$ + 10
= −2\$ − 3.5 + 24.5 − 20
0 = \$ − 2\$ − 5
= −2\$ − 3.5 + 4.5
\$−2=0
\$−5=0
\$=2
\$=5
ℎ=
2+5
= 3.5
2
, = −23.5 + 143.5 − 20 = 4.5
The maximum profit is \$450 000.
#10.
Method 1: (Partial Factor)
Method 2: (complete square)
Let ℎ = \$
ℎ = 65 + 10 − 5 65 = 65 + 10 − 5 ℎ = −5 + 10 + 65
0 = −5 + 10
ℎ = −5 − 2 + 1 − 1 + 65
−5 − 2 = 0
ℎ = −5 − 2 + 1 − 5−1 + 65
−5 = 0
−2=0
ℎ = −5 − 1 + 5 + 65
=0
=2
ℎ = −5 − 1 + 70
ℎ=
0+2
=1
2
The ball reaches the maximum height at 1 second.
The ball reaches the maximum height at 1 second.