H Algebra 2/Trig · Unit 2
Find a starting height, a vertex, or when something hits the ground. The vertex is also a maximum area or a maximum profit.
Jump to section:
t = −b2a
start: t = 0 · ground: h = 0
This is the same as the y-intercept.
Example
A ball is thrown upward from a platform. Its height is h(t) = −16t² + 40t + 24, where t is the time in seconds and h(t) is the height in feet.What is the starting height?
The starting height is when t = 0.
h(t) = −16t² + 40t + 24
h(0) = −16(0)² + 40(0) + 24
h(0) = 0 + 0 + 24
h(0) = 24
starting height = 24 feet
Set the equation = 0 and solve by factoring or using the quadratic formula.
Example
A ball is thrown upward. Its height is h(t) = −16t² + 48t + 64, where t is the time in seconds and h(t) is the height in feet.When does the ball hit the ground?
Set h(t) = 0.
−16t² + 48t + 64 = 0
Divide by −16.
t² − 3t − 4 = 0
(t + 1)(t − 4) = 0
Set each factor equal to zero.
Time cannot be negative.
hits the ground = 4 seconds
The x-value is the time of the maximum. The y-value is the maximum value.
Example
A ball is thrown upward. Its height is h(t) = −16t² + 32t + 48, where t is the time in seconds and h(t) is the height in feet.When does the ball reach its highest point, and what is that height?
The formula is
t = −b2a
a = −16, b = 32
Substitute.
t = −32−32
t = 3232
t = 1
Substitute t = 1 into h(t).
h(1) = −16(1)² + 32(1) + 48
h(1) = −16(1) + 32(1) + 48
h(1) = −16 + 32 + 48
h(1) = 64
time for max height = 1 second
max height = 64 feet
The vertex is (1 second, 64 feet).
Substitute t = 0. That is the y-intercept.
The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 48t + 18
h(0) = −16(0)² + 48(0) + 18
h(0) = 0 + 0 + 18
h(0) = 18
Answer: starting height = 18 feet
The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 64t + 5
h(0) = −16(0)² + 64(0) + 5
h(0) = 0 + 0 + 5
h(0) = 5
Answer: starting height = 5 feet
The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 32t + 36
h(0) = −16(0)² + 32(0) + 36
h(0) = 0 + 0 + 36
h(0) = 36
Answer: starting height = 36 feet
The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 80t + 9
h(0) = −16(0)² + 80(0) + 9
h(0) = 0 + 0 + 9
h(0) = 9
Answer: starting height = 9 feet
The starting height is when t = 0. This is the y-intercept.
h(t) = 12 + 24t − 16t²
h(0) = 12 + 24(0) − 16(0)²
h(0) = 12 + 0 − 0
h(0) = 12
Answer: starting height = 12 feet
Use the vertex formula.
The formula is
t = −b2a
a = −16, b = 64
Substitute.
t = −64−32
t = 6432
t = 2
Answer: time for max height = 2 seconds
The formula is
t = −b2a
a = −16, b = 32
Substitute.
t = −32−32
t = 3232
t = 1
Answer: time for max height = 1 second
The formula is
t = −b2a
a = −16, b = 96
Substitute.
t = −96−32
t = 9632
t = 3
Answer: time for max height = 3 seconds
The formula is
t = −b2a
a = −16, b = 16
Substitute.
t = −16−32
t = 1632
t = 12
Answer: time for max height = 12 second
The formula is
t = −b2a
a = −16, b = 48
Substitute.
t = −48−32
t = 4832
t = 32
Answer: time for max height = 32 seconds
Find the vertex time, then substitute.
The formula is
t = −b2a
a = −16, b = 32
Substitute.
t = −32−32
t = 3232
t = 1
Substitute t = 1 into h(t).
h(1) = −16(1)² + 32(1) + 21
h(1) = −16(1) + 32(1) + 21
h(1) = −16 + 32 + 21
h(1) = 37
Answer: max height = 37 feet
The formula is
t = −b2a
a = −16, b = 64
Substitute.
t = −64−32
t = 6432
t = 2
Substitute t = 2 into h(t).
h(2) = −16(2)² + 64(2) + 3
h(2) = −16(4) + 64(2) + 3
h(2) = −64 + 128 + 3
h(2) = 67
Answer: max height = 67 feet
The formula is
t = −b2a
a = −16, b = 96
Substitute.
t = −96−32
t = 9632
t = 3
Substitute t = 3 into h(t).
h(3) = −16(3)² + 96(3) + 15
h(3) = −16(9) + 96(3) + 15
h(3) = −144 + 288 + 15
h(3) = 159
Answer: max height = 159 feet
The formula is
t = −b2a
a = −16, b = 16
Substitute.
t = −16−32
t = 1632
t = 12
Substitute t = 12 into h(t).
h(12) = −16(12)² + 16(12) + 11
(12)² = 14
−16 · 14 = −4
16 · 12 = 8
h(12) = −4 + 8 + 11
h(12) = 15
Answer: max height = 15 feet
The formula is
t = −b2a
a = −16, b = 48
Substitute.
t = −48−32
t = 4832
t = 32
Substitute t = 32 into h(t).
h(32) = −16(32)² + 48(32) + 7
(32)² = 94
−16 · 94 = −36
48 · 32 = 72
h(32) = −36 + 72 + 7
h(32) = 43
Answer: max height = 43 feet
Set the height equal to 0 and factor. The horizontal axis may be time or distance from the start.
Set h(t) = 0.
−16t² + 16t + 32 = 0
Divide by −16.
t² − t − 2 = 0
(t + 1)(t − 2) = 0
Set each factor equal to zero.
Time cannot be negative.
Answer: hits the ground = 2 seconds
Set h(t) = 0.
−16t² + 96t + 112 = 0
Divide by −16.
t² − 6t − 7 = 0
(t + 1)(t − 7) = 0
Set each factor equal to zero.
Time cannot be negative.
Answer: hits the ground = 7 seconds
Set h(t) = 0.
−16t² + 80t = 0
Divide by −16.
t² − 5t = 0
t(t − 5) = 0
Set each factor equal to zero.
t = 0 is the start.
Answer: hits the ground = 5 seconds
Set h(x) = 0.
−x² + 6x + 16 = 0
Multiply both sides by −1.
x² − 6x − 16 = 0
(x + 2)(x − 8) = 0
Set each factor equal to zero.
Distance from the start cannot be negative.
Answer: hits the ground = 8 yards from the start
Set h(x) = 0.
−x² + 4x + 5 = 0
Multiply both sides by −1.
x² − 4x − 5 = 0
(x + 1)(x − 5) = 0
Set each factor equal to zero.
Distance from the start cannot be negative.
Answer: hits the ground = 5 meters from the start
The height is not the maximum and not the ground.
Set h(t) = 68.
−16t² + 64t + 20 = 68
Subtract 68 from both sides.
−16t² + 64t − 48 = 0
Divide by −16.
t² − 4t + 3 = 0
(t − 1)(t − 3) = 0
Set each factor equal to zero.
Answer: 68 feet high at 1 second and at 3 seconds
Set h(t) = 140.
−16t² + 96t + 12 = 140
Subtract 140 from both sides.
−16t² + 96t − 128 = 0
Divide by −16.
t² − 6t + 8 = 0
(t − 2)(t − 4) = 0
Set each factor equal to zero.
Answer: 140 feet high at 2 seconds and at 4 seconds
Set h(t) = 60.
−16t² + 48t + 28 = 60
Subtract 60 from both sides.
−16t² + 48t − 32 = 0
Divide by −16.
t² − 3t + 2 = 0
(t − 1)(t − 2) = 0
Set each factor equal to zero.
Answer: 60 feet high at 1 second and at 2 seconds
Set h(t) = 89.
−16t² + 80t + 25 = 89
Subtract 89 from both sides.
−16t² + 80t − 64 = 0
Divide by −16.
t² − 5t + 4 = 0
(t − 1)(t − 4) = 0
Set each factor equal to zero.
Answer: 89 feet high at 1 second and at 4 seconds
Set h(t) = 40.
−5t² + 20t + 25 = 40
Subtract 40 from both sides.
−5t² + 20t − 15 = 0
Divide by −5.
t² − 4t + 3 = 0
(t − 1)(t − 3) = 0
Set each factor equal to zero.
Answer: 40 meters high at 1 second and at 3 seconds
The vertex is the maximum area.
a. Area = (length)(width).
A = x(50 − x)
A = 50x − x²
b. The vertex is the maximum area.
The formula is
x = −b2a
a = −1, b = 50
Substitute.
x = −50−2
x = 502
x = 25
c. Substitute x = 25.
A = 25(50 − 25)
A = 25(25)
A = 625
Answer: area = 50x − x²width = 25 feetmaximum area = 625 square feet
a. Area = (length)(width).
A = x(28 − x)
A = 28x − x²
b. The vertex is the maximum area.
The formula is
x = −b2a
a = −1, b = 28
Substitute.
x = −28−2
x = 282
x = 14
c. Substitute x = 14.
A = 14(28 − 14)
A = 14(14)
A = 196
Answer: area = 28x − x²width = 14 metersmaximum area = 196 square meters
The side opposite the barn is 60 − 2x.
A = x(60 − 2x)
A = 60x − 2x²
b. The vertex is the maximum area.
The formula is
x = −b2a
a = −2, b = 60
Substitute.
x = −60−4
x = 604
x = 15
The side opposite the barn is
60 − 2(15)
60 − 30
30 feet
c. Substitute.
A = 15(30)
A = 450
Answer: area = 60x − 2x²width = 15 feetmaximum area = 450 square feet
Area = (length)(width).
A = x(64 − x)
A = 64x − x²
The vertex is the maximum area.
The formula is
x = −b2a
a = −1, b = 64
Substitute.
x = −64−2
x = 642
x = 32
Substitute x = 32.
A = 32(64 − 32)
A = 32(32)
A = 1,024
Answer: width = 32 feetmaximum area = 1,024 square feet
The side opposite the wall is 48 − 2x.
A = x(48 − 2x)
A = 48x − 2x²
The vertex is the maximum area.
The formula is
x = −b2a
a = −2, b = 48
Substitute.
x = −48−4
x = 484
x = 12
The side opposite the wall is
48 − 2(12)
48 − 24
24 feet
Substitute.
A = 12(24)
A = 288
Answer: width = 12 feetmaximum area = 288 square feet
The vertex is the maximum profit.
P(x) = −2x² + 40x − 48
The vertex is the maximum profit.
The formula is
x = −b2a
a = −2, b = 40
Substitute.
x = −40−4
x = 404
x = 10
Substitute x = 10 into P(x).
P(10) = −2(10)² + 40(10) − 48
P(10) = −2(100) + 40(10) − 48
P(10) = −200 + 400 − 48
P(10) = 152
Answer: number of candles = 10maximum profit = $152
P(n) = −n² + 30n − 56
The vertex is the maximum profit.
The formula is
n = −b2a
a = −1, b = 30
Substitute.
n = −30−2
n = 302
n = 15
Substitute n = 15 into P(n).
P(15) = −(15)² + 30(15) − 56
P(15) = −(225) + 30(15) − 56
P(15) = −225 + 450 − 56
P(15) = 169
Answer: number of pies = 15maximum profit = $169
P(h) = −3h² + 48h − 60
The vertex is the maximum profit.
The formula is
h = −b2a
a = −3, b = 48
Substitute.
h = −48−6
h = 486
h = 8
Substitute h = 8 into P(h).
P(8) = −3(8)² + 48(8) − 60
P(8) = −3(64) + 48(8) − 60
P(8) = −192 + 384 − 60
P(8) = 132
Answer: number of hours = 8maximum profit = $132
Profit = (price − cost)(number sold).
P(x) = (x − 8)(50 − x)
Expand.
P(x) = 50x − x² − 400 + 8x
P(x) = −x² + 58x − 400
The vertex is the maximum profit.
The formula is
x = −b2a
a = −1, b = 58
Substitute.
x = −58−2
x = 582
x = 29
Substitute x = 29 into P(x) = (x − 8)(50 − x).
P(29) = (29 − 8)(50 − 29)
P(29) = (21)(21)
P(29) = 441
Answer: price = $29maximum profit = $441
Profit = (price − cost)(number sold).
P(x) = (x − 2)(20 − x)
Expand.
P(x) = 20x − x² − 40 + 2x
P(x) = −x² + 22x − 40
The vertex is the maximum profit.
The formula is
x = −b2a
a = −1, b = 22
Substitute.
x = −22−2
x = 222
x = 11
Substitute x = 11 into P(x) = (x − 2)(20 − x).
P(11) = (11 − 2)(20 − 11)
P(11) = (9)(9)
P(11) = 81
Answer: price = $11maximum profit = $81
A full projectile, a maximum area, or a maximum profit.
a. The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 40t + 9
h(0) = −16(0)² + 40(0) + 9
h(0) = 0 + 0 + 9
h(0) = 9
b. The ball is highest at the vertex.
The formula is
t = −b2a
a = −16, b = 40
Substitute.
t = −40−32
t = 4032
t = 54
c. Substitute t = 54 into h(t).
h(54) = −16(54)² + 40(54) + 9
(54)² = 2516
−16 · 2516 = −25
40 · 54 = 50
h(54) = −25 + 50 + 9
h(54) = 34
d. The ball hits the ground when h(t) = 0.
−16t² + 40t + 9 = 0
The formula is
t = −b ± √(b² − 4ac)2a
t = −40 ± √(1600 + 576)−32
t = −40 ± √2176−32
Use the positive time.
t = 40 + √217632
t ≈ 2.71 seconds
Answer: starting height = 9 feettime for max height = 54 secondsmax height = 34 feethits the ground ≈ 2.71 seconds
a. The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 48t + 5
h(0) = −16(0)² + 48(0) + 5
h(0) = 0 + 0 + 5
h(0) = 5
b. The ball is highest at the vertex.
The formula is
t = −b2a
a = −16, b = 48
Substitute.
t = −48−32
t = 4832
t = 32
c. Substitute t = 32 into h(t).
h(32) = −16(32)² + 48(32) + 5
(32)² = 94
−16 · 94 = −36
48 · 32 = 72
h(32) = −36 + 72 + 5
h(32) = 41
d. The ball hits the ground when h(t) = 0.
−16t² + 48t + 5 = 0
The formula is
t = −b ± √(b² − 4ac)2a
t = −48 ± √(2304 + 320)−32
t = 48 + √262432
t ≈ 3.10 seconds
Answer: starting height = 5 feettime for max height = 32 secondsmax height = 41 feethits the ground ≈ 3.10 seconds
a. The starting height is when t = 0. This is the y-intercept.
h(t) = −16t² + 72t
h(0) = −16(0)² + 72(0)
h(0) = 0 + 0
h(0) = 0
b. The ball is highest at the vertex.
The formula is
t = −b2a
a = −16, b = 72
Substitute.
t = −72−32
t = 7232
t = 94
c. Substitute t = 94 into h(t).
h(94) = −16(94)² + 72(94)
(94)² = 8116
−16 · 8116 = −81
72 · 94 = 162
h(94) = −81 + 162
h(94) = 81
d. The ball hits the ground when h(t) = 0.
−16t² + 72t = 0
The GCF is −8t.
−8t(2t − 9) = 0
Set each factor equal to zero.
t = 0 is the start.
Answer: starting height = 0 feettime for max height = 94 secondsmax height = 81 feethits the ground = 92 seconds
a. Area = (length)(width).
A = x(70 − x)
A = 70x − x²
b. The vertex is the maximum area.
The formula is
x = −b2a
a = −1, b = 70
Substitute.
x = −70−2
x = 702
x = 35
c. Substitute x = 35.
A = 35(70 − 35)
A = 35(35)
A = 1,225
Answer: area = 70x − x²width = 35 metersmaximum area = 1,225 square meters
a. Area = (length)(width).
A = x(90 − x)
A = 90x − x²
b. The vertex is the maximum area.
The formula is
x = −b2a
a = −1, b = 90
Substitute.
x = −90−2
x = 902
x = 45
c. Substitute x = 45.
A = 45(90 − 45)
A = 45(45)
A = 2,025
Answer: area = 90x − x²width = 45 feetmaximum area = 2,025 square feet
N = 64d − d²
The vertex is the maximum number sold.
The formula is
d = −b2a
a = −1, b = 64
Substitute.
d = −64−2
d = 642
d = 32
Substitute d = 32.
N = 64(32) − (32)²
N = 2048 − 1024
N = 1024
Answer: price = $32hats sold = 1,024
P(x) = −4x² + 120x − 500
The vertex is the maximum profit.
The formula is
x = −b2a
a = −4, b = 120
Substitute.
x = −120−8
x = 1208
x = 15
Substitute x = 15 into P(x).
P(15) = −4(15)² + 120(15) − 500
P(15) = −4(225) + 120(15) − 500
P(15) = −900 + 1800 − 500
P(15) = 400
Answer: number of cakes = 15maximum profit = $400
Let n be the smaller positive integer. Write an equation, then solve by factoring.
Let n be the smaller positive odd integer.
The next consecutive odd integer is n + 2.
n² + (n + 2)² = 394
(n + 2)² = n² + 4n + 4
Substitute.
n² + (n² + 4n + 4) = 394
2n² + 4n + 4 = 394
Subtract 394 from both sides.
2n² + 4n + 4 − 394 = 0
2n² + 4n − 390 = 0
Divide both sides by 2.
n² + 2n − 195 = 0
Find two numbers that multiply to −195 and add to 2.
15 and −13.
15 · (−13) = −195
15 + (−13) = 2
(n + 15)(n − 13) = 0
Set each factor equal to zero.
n = −15 is not positive.
The next consecutive odd integer is n + 2.
n + 2 = 13 + 2
n + 2 = 15
Check.
13² = 169
15² = 225
169 + 225 = 394
Answer: the two integers are 13 and 15
Let n be the smaller positive odd integer.
The next consecutive odd integer is n + 2.
n² + (n + 2)² = 802
(n + 2)² = n² + 4n + 4
Substitute.
n² + (n² + 4n + 4) = 802
2n² + 4n + 4 = 802
Subtract 802 from both sides.
2n² + 4n + 4 − 802 = 0
2n² + 4n − 798 = 0
Divide both sides by 2.
n² + 2n − 399 = 0
Find two numbers that multiply to −399 and add to 2.
21 and −19.
21 · (−19) = −399
21 + (−19) = 2
(n + 21)(n − 19) = 0
Set each factor equal to zero.
n = −21 is not positive.
The next consecutive odd integer is n + 2.
n + 2 = 19 + 2
n + 2 = 21
Check.
19² = 361
21² = 441
361 + 441 = 802
Answer: the two integers are 19 and 21
Let n be the smaller positive integer.
The other integer is n + 6.
n(n + 6) + n + (n + 6) = 111
n(n + 6) = n² + 6n
n + (n + 6) = 2n + 6
Substitute.
n² + 6n + 2n + 6 = 111
n² + 8n + 6 = 111
Subtract 111 from both sides.
n² + 8n + 6 − 111 = 0
n² + 8n − 105 = 0
Find two numbers that multiply to −105 and add to 8.
15 and −7.
15 · (−7) = −105
15 + (−7) = 8
(n + 15)(n − 7) = 0
Set each factor equal to zero.
n = −15 is not positive.
The other integer is n + 6.
n + 6 = 7 + 6
n + 6 = 13
Check.
7 · 13 = 91
7 + 13 = 20
91 + 20 = 111
Answer: the two integers are 7 and 13
Let n be the smaller positive integer.
The other integer is n + 4.
n(n + 4) + n + (n + 4) = 116
n(n + 4) = n² + 4n
n + (n + 4) = 2n + 4
Substitute.
n² + 4n + 2n + 4 = 116
n² + 6n + 4 = 116
Subtract 116 from both sides.
n² + 6n + 4 − 116 = 0
n² + 6n − 112 = 0
Find two numbers that multiply to −112 and add to 6.
14 and −8.
14 · (−8) = −112
14 + (−8) = 6
(n + 14)(n − 8) = 0
Set each factor equal to zero.
n = −14 is not positive.
The other integer is n + 4.
n + 4 = 8 + 4
n + 4 = 12
Check.
8 · 12 = 96
8 + 12 = 20
96 + 20 = 116
Answer: the two integers are 8 and 12
Let n be the smaller positive integer.
The other integer is n + 9.
n(n + 9) + n + (n + 9) = 219
n(n + 9) = n² + 9n
n + (n + 9) = 2n + 9
Substitute.
n² + 9n + 2n + 9 = 219
n² + 11n + 9 = 219
Subtract 219 from both sides.
n² + 11n + 9 − 219 = 0
n² + 11n − 210 = 0
Find two numbers that multiply to −210 and add to 11.
21 and −10.
21 · (−10) = −210
21 + (−10) = 11
(n + 21)(n − 10) = 0
Set each factor equal to zero.
n = −21 is not positive.
The other integer is n + 9.
n + 9 = 10 + 9
n + 9 = 19
Check.
10 · 19 = 190
10 + 19 = 29
190 + 29 = 219
Answer: the two integers are 10 and 19
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