H Algebra 2/Trig · Unit 2
Match the information to a method. Three points use quadratic regression. Two zeros use root form. A given equation is ready to solve. A vertex plus a point uses vertex form.
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3 points: QuadReg
2 zeros: root form
equation given: solve it
vertex given: vertex form
Three points
Use quadratic regression when the problem gives three points and asks for the quadratic through them.
On a TI-84 Plus CE: Stat Edit, Stat Calc, 5: Quad Regression. The calculator writes standard form y = ax² + bx + c.
Two zeros
Use root form when the output is 0 at two inputs and you know one other point.
The formula is y = a(x − r)(x − s). The other point finds a.
Equation given, height
When a projectile equation is already given, solve that equation.
A squared time with no single t term calls for the square-root method. Keep the positive time.
Equation given, maximum
When a standard-form equation opens downward, the maximum is the vertex.
The formula is x = −b2a. Substitute that input to find the maximum value.
Vertex given
Use vertex form when the problem gives a maximum or a minimum and one other point.
The formula is y = a(x − h)² + k. The other point finds a.
Three points are given. Use QuadReg on a TI-84 Plus CE, then answer the question.
Quadratic regression is a good fit. The problem gives three roster counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the number of years since 2014.
y is the dependent variable: the number of members.
The points are (2, 40), (5, 55), (9, 47).
| L1 | L2 |
|---|---|
| 2 | 40 |
| 5 | 55 |
| 9 | 47 |
The years 2016, 2019, and 2023 are not entered into L1. x is the number of years since 2014, so L1 gets 2, 5, and 9.
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 2 | 40 | |
| 2 | 5 | 55 | |
| 3 | 9 | 47 | |
| 4 |
Stat Calc
5: Quad Regression
a = −1, b = 12, and c = 20.
If a value is a tiny bit off a whole number, such as −1.000000001, write −1.
The formula is
y = −x² + 12x + 20
2020 is 6 years after 2014.
y(6) = −(6)² + 12(6) + 20
y(6) = −36 + 72 + 20
y(6) = 56
Answer: x is the independent variable: the number of years since 2014.y is the dependent variable: the number of members.y = −x² + 12x + 20The model predicts 56 members.
Quadratic regression is a good fit. The problem gives three enrollment counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the number of years since 2010.
y is the dependent variable: the number of students.
The points are (1, 48), (4, 78), (8, 62).
| L1 | L2 |
|---|---|
| 1 | 48 |
| 4 | 78 |
| 8 | 62 |
The years 2011, 2014, and 2018 are not entered into L1. x is the number of years since 2010, so L1 gets 1, 4, and 8.
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 1 | 48 | |
| 2 | 4 | 78 | |
| 3 | 8 | 62 | |
| 4 |
Stat Calc
5: Quad Regression
a = −2, b = 20, and c = 30.
The formula is
y = −2x² + 20x + 30
2016 is 6 years after 2010.
y(6) = −2(6)² + 20(6) + 30
y(6) = −72 + 120 + 30
y(6) = 78
Answer: x is the independent variable: the number of years since 2010.y is the dependent variable: the number of students.y = −2x² + 20x + 30The model predicts 78 students.
Quadratic regression is a good fit. The problem gives three choir counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the number of years since 2012.
y is the dependent variable: the number of singers.
The points are (1, 31), (3, 39), (7, 31).
| L1 | L2 |
|---|---|
| 1 | 31 |
| 3 | 39 |
| 7 | 31 |
The years 2013, 2015, and 2019 are not entered into L1. x is the number of years since 2012, so L1 gets 1, 3, and 7.
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 1 | 31 | |
| 2 | 3 | 39 | |
| 3 | 7 | 31 | |
| 4 |
Stat Calc
5: Quad Regression
a = −1, b = 8, and c = 24.
The formula is
y = −x² + 8x + 24
2017 is 5 years after 2012.
y(5) = −(5)² + 8(5) + 24
y(5) = −25 + 40 + 24
y(5) = 39
Answer: x is the independent variable: the number of years since 2012.y is the dependent variable: the number of singers.y = −x² + 8x + 24The model predicts 39 singers.
Quadratic regression is a good fit. The problem gives three project counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the number of years since 2016.
y is the dependent variable: the number of projects.
The points are (1, 36), (3, 60), (6, 51).
| L1 | L2 |
|---|---|
| 1 | 36 |
| 3 | 60 |
| 6 | 51 |
The years 2017, 2019, and 2022 are not entered into L1. x is the number of years since 2016, so L1 gets 1, 3, and 6.
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 1 | 36 | |
| 2 | 3 | 60 | |
| 3 | 6 | 51 | |
| 4 |
Stat Calc
5: Quad Regression
a = −3, b = 24, and c = 15.
The formula is
y = −3x² + 24x + 15
2020 is 4 years after 2016.
y(4) = −3(4)² + 24(4) + 15
y(4) = −48 + 96 + 15
y(4) = 63
Answer: x is the independent variable: the number of years since 2016.y is the dependent variable: the number of projects.y = −3x² + 24x + 15The model predicts 63 projects.
Quadratic regression is a good fit. The problem gives three sales counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the week number.
y is the dependent variable: the number of cups.
The points are (1, 54), (3, 70), (6, 64).
| L1 | L2 |
|---|---|
| 1 | 54 |
| 3 | 70 |
| 6 | 64 |
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 1 | 54 | |
| 2 | 3 | 70 | |
| 3 | 6 | 64 | |
| 4 |
Stat Calc
5: Quad Regression
a = −2, b = 16, and c = 40.
The formula is
y = −2x² + 16x + 40
Week 4 means x = 4.
y(4) = −2(4)² + 16(4) + 40
y(4) = −32 + 64 + 40
y(4) = 72
Answer: x is the independent variable: the week number.y is the dependent variable: the number of cups.y = −2x² + 16x + 40The model predicts 72 cups.
Quadratic regression is a good fit. The problem gives three salmon counts, and QuadReg finds the one quadratic through those points.
x is the independent variable: the number of years since the project.
y is the dependent variable: the number of salmon.
The points are (2, 56), (5, 65), (8, 56).
| L1 | L2 |
|---|---|
| 2 | 56 |
| 5 | 65 |
| 8 | 56 |
Stat Edit
| L1 | L2 | L3 | |
|---|---|---|---|
| 1 | 2 | 56 | |
| 2 | 5 | 65 | |
| 3 | 8 | 56 | |
| 4 |
Stat Calc
5: Quad Regression
a = −1, b = 10, and c = 40.
The formula is
y = −x² + 10x + 40
Year 4 means x = 4.
y(4) = −(4)² + 10(4) + 40
y(4) = −16 + 40 + 40
y(4) = 64
Answer: x is the independent variable: the number of years since the project.y is the dependent variable: the number of salmon.y = −x² + 10x + 40The model predicts 64 salmon.
The value is 0 at two inputs. Write root form, use the other point to find a, then answer the question.
Root form is a good fit. The rocket is on the ground at 0 seconds and again at 8 seconds, so those times are the roots. The height at 2 seconds finds a.
The formula is
h(t) = a(t − r)(t − s)
The zeros are t = 0 and t = 8.
h(t) = a(t − 0)(t − 8)
Substitute the known point, t = 2 and h = 48.
48 = a(2 − 0)(2 − 8)
48 = a(2)(−6)
48 = −12a
a = −4812
a = −4
h(t) = −4(t − 0)(t − 8)
Substitute t = 5.
h(5) = −4(5 − 0)(5 − 8)
h(5) = −4(5)(−3)
h(5) = 60
Answer: h(t) = −4(t − 0)(t − 8)The rocket is 60 feet high after 5 seconds.
Root form is a good fit. The ball is on the ground at 0 seconds and again at 4 seconds, so those times are the roots. The height at 1 second finds a.
The formula is
h(t) = a(t − r)(t − s)
The zeros are t = 0 and t = 4.
h(t) = a(t − 0)(t − 4)
Substitute the known point, t = 1 and h = 30.
30 = a(1 − 0)(1 − 4)
30 = a(1)(−3)
30 = −3a
a = −303
a = −10
h(t) = −10(t − 0)(t − 4)
Substitute t = 2.
h(2) = −10(2 − 0)(2 − 4)
h(2) = −10(2)(−2)
h(2) = 40
Answer: h(t) = −10(t − 0)(t − 4)The ball is 40 feet high after 2 seconds.
Root form is a good fit. The ball is on the ground at 0 seconds and again at 6 seconds, so those times are the roots. The height at 2 seconds finds a.
The formula is
h(t) = a(t − r)(t − s)
The zeros are t = 0 and t = 6.
h(t) = a(t − 0)(t − 6)
Substitute the known point, t = 2 and h = 48.
48 = a(2 − 0)(2 − 6)
48 = a(2)(−4)
48 = −8a
a = −488
a = −6
h(t) = −6(t − 0)(t − 6)
Substitute t = 3.
h(3) = −6(3 − 0)(3 − 6)
h(3) = −6(3)(−3)
h(3) = 54
Answer: h(t) = −6(t − 0)(t − 6)The ball is 54 feet high after 3 seconds.
Root form is a good fit. The firework is on the ground at 0 seconds and again at 10 seconds, so those times are the roots. The height at 2 seconds finds a.
The formula is
h(t) = a(t − r)(t − s)
The zeros are t = 0 and t = 10.
h(t) = a(t − 0)(t − 10)
Substitute the known point, t = 2 and h = 80.
80 = a(2 − 0)(2 − 10)
80 = a(2)(−8)
80 = −16a
a = −8016
a = −5
h(t) = −5(t − 0)(t − 10)
Substitute t = 6.
h(6) = −5(6 − 0)(6 − 10)
h(6) = −5(6)(−4)
h(6) = 120
Answer: h(t) = −5(t − 0)(t − 10)The firework is 120 feet high after 6 seconds.
Root form is a good fit. The height is 0 at both ends, x = 0 and x = 12, so those distances are the roots. The height 4 feet from the left end finds a.
The formula is
h(x) = a(x − r)(x − s)
The zeros are x = 0 and x = 12.
h(x) = a(x − 0)(x − 12)
Substitute the known point, x = 4 and h = 16.
16 = a(4 − 0)(4 − 12)
16 = a(4)(−8)
16 = −32a
a = −1632
a = −12
h(x) = −12(x − 0)(x − 12)
Substitute x = 6.
h(6) = −12(6 − 0)(6 − 12)
h(6) = −12(6)(−6)
h(6) = 12·36
h(6) = 362
h(6) = 18
Answer: h(x) = −12(x − 0)(x − 12)The arch is 18 feet high.
Root form is a good fit. The height is 0 at x = 2 and at x = 8, so those distances are the roots. The height at 4 feet finds a.
The formula is
h(x) = a(x − r)(x − s)
The zeros are x = 2 and x = 8.
h(x) = a(x − 2)(x − 8)
Substitute the known point, x = 4 and h = 12.
12 = a(4 − 2)(4 − 8)
12 = a(2)(−4)
12 = −8a
a = −128
a = −32
h(x) = −32(x − 2)(x − 8)
Substitute x = 5.
h(5) = −32(5 − 2)(5 − 8)
h(5) = −32(3)(−3)
h(5) = 32·9
h(5) = 272
Answer: h(x) = −32(x − 2)(x − 8)The stream is 272 feet high.
The height equation is given. Set the height equal to the number in the question and solve.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The ball hits the water when the height is 0.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 144). The graph opens downward.
not drawn to scale
The formula is
h(t) = 144 − 16t²
The ball hits the water when the height is 0.
0 = 144 − 16t²
16t² = 144
t² = 14416
t² = 9
t = ±
t = ± 3
Time is positive.
t = 3
Answer: The ball hits the water after 3 seconds.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The rock hits the ground when the height is 0.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 64). The graph opens downward.
not drawn to scale
The formula is
h(t) = 64 − 16t²
The rock hits the ground when the height is 0.
0 = 64 − 16t²
16t² = 64
t² = 6416
t² = 4
t = ±
t = ± 2
Time is positive.
t = 2
Answer: The rock hits the ground after 2 seconds.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The balloon hits the ground when the height is 0.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 80). The graph opens downward.
not drawn to scale
The formula is
h(t) = 80 − 16t²
The balloon hits the ground when the height is 0.
0 = 80 − 16t²
16t² = 80
t² = 8016
t² = 5
t = ±
Time is positive.
t =
Answer: The balloon hits the ground after seconds.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The pumpkin hits the ground when the height is 0.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 196). The graph opens downward.
not drawn to scale
The formula is
h(t) = 196 − 16t²
The pumpkin hits the ground when the height is 0.
0 = 196 − 16t²
16t² = 196
t² = 19616
t² = 494
t = ±
t = ± 72
Time is positive.
t = 72
Answer: The pumpkin hits the ground after 72 seconds.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The stick is at the asked height when h(t) = 112.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 256). The graph opens downward.
not drawn to scale
The formula is
h(t) = 256 − 16t²
The stick is 112 feet above the ground when the height is 112.
112 = 256 − 16t²
16t² = 256 − 112
16t² = 144
t² = 14416
t² = 9
t = ±
t = ± 3
Time is positive.
t = 3
Answer: The stick is 112 feet above the ground after 3 seconds.
The square-root method is a good fit. The height equation is given, and it has a squared time with no single t term. The platform is at ground level when the height is 0.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 75). The graph opens downward.
not drawn to scale
The formula is
h(t) = 75 − 3t²
The platform is at ground level when the height is 0.
0 = 75 − 3t²
3t² = 75
t² = 753
t² = 25
t = ±
t = ± 5
Time is positive.
t = 5
Answer: The platform is at ground level after 5 seconds.
The equation is given in standard form, and a is negative. The maximum is the vertex.
The vertex formula is a good fit. The profit equation is in standard form, and a is negative, so the vertex is the maximum profit.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, −14). The graph opens downward.
not drawn to scale
p(x) = −2x² + 16x − 14
The formula is
x = −b2a
a = −2 and b = 16.
Substitute.
x = −16−4
x = 164
x = 4
x counts thousands of headphones.
number = 4 × 1,000
number = 4,000
Substitute x = 4.
p(4) = −2(4)² + 16(4) − 14
p(4) = −32 + 64 − 14
p(4) = 18
Answer: 4,000 headphones18 million dollars
The vertex formula is a good fit. The profit equation is in standard form, and a is negative, so the vertex is the maximum profit.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, −20). The graph opens downward.
not drawn to scale
p(x) = −x² + 12x − 20
The formula is
x = −b2a
a = −1 and b = 12.
Substitute.
x = −12−2
x = 122
x = 6
x counts thousands of cases.
number = 6 × 1,000
number = 6,000
Substitute x = 6.
p(6) = −(6)² + 12(6) − 20
p(6) = −36 + 72 − 20
p(6) = 16
Answer: 6,000 cases16 million dollars
The vertex formula is a good fit. The profit equation is in standard form, and a is negative, so the vertex is the maximum profit.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, −15). The graph opens downward.
not drawn to scale
p(x) = −3x² + 18x − 15
The formula is
x = −b2a
a = −3 and b = 18.
Substitute.
x = −18−6
x = 186
x = 3
x counts thousands of bottles.
number = 3 × 1,000
number = 3,000
Substitute x = 3.
p(3) = −3(3)² + 18(3) − 15
p(3) = −27 + 54 − 15
p(3) = 12
Answer: 3,000 bottles12 million dollars
The vertex formula is a good fit. The profit equation is in standard form, and a is negative, so the vertex is the maximum profit.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, −5). The graph opens downward.
not drawn to scale
p(x) = −4x² + 12x − 5
The formula is
x = −b2a
a = −4 and b = 12.
Substitute.
x = −12−8
x = 128
x = 32
x counts thousands of jackets.
number = 32 × 1,000
number = 30002
number = 1,500
Substitute x = 32.
p(32) = −432² + 1232 − 5
32² = 94
−4 · 94 = −9
12 · 32 = 18
p(32) = −9 + 18 − 5
p(32) = 4
Answer: 1,500 jackets4 million dollars
The vertex formula is a good fit. The profit equation is in standard form, and a is negative, so the vertex is the maximum profit.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, −1500). The graph opens downward.
not drawn to scale
P(n) = −5n² + 400n − 1500
The formula is
n = −b2a
a = −5 and b = 400.
Substitute.
n = −400−10
n = 40010
n = 40
n is the ticket price in dollars.
Substitute n = 40.
P(40) = −5(40)² + 400(40) − 1500
P(40) = −8000 + 16000 − 1500
P(40) = 6500
Answer: ticket price = 40 dollarsmaximum profit = 6,500 dollars
The vertex formula is a good fit. The yield equation is in standard form, and a is negative, so the vertex is the greatest yield.
The y-intercept is easy to graph, and you know the direction of the graph.
The y-intercept is (0, 6). The graph opens downward.
not drawn to scale
y(x) = −2x² + 24x + 6
The formula is
x = −b2a
a = −2 and b = 24.
Substitute.
x = −24−4
x = 244
x = 6
Substitute x = 6.
y(6) = −2(6)² + 24(6) + 6
y(6) = −72 + 144 + 6
y(6) = 78
Answer: 6 plants in the row78 tomatoes on each plant
A maximum or a minimum is given, along with one other point. Write vertex form, find a, then answer the question.
Vertex form is a good fit. The greatest height is 40 cm, and it happens at a spacing of 12 cm, so (12, 40) is the vertex. The height at 15 cm finds a.
The formula is
y = a(x − h)² + k
The vertex is (12, 40).
y(x) = a(x − 12)² + 40
Substitute (15, 31).
31 = a(15 − 12)² + 40
31 = a(3)² + 40
31 = 9a + 40
31 − 40 = 9a
−9 = 9a
a = −99
a = −1
y(x) = −(x − 12)² + 40
Substitute x = 10.
y(10) = −(10 − 12)² + 40
y(10) = −(−2)² + 40
(−2)² = 4
y(10) = −4 + 40
y(10) = 36
Answer: y(x) = −(x − 12)² + 40The plants are 36 cm tall.
Vertex form is a good fit. The least height is 18 inches, and it happens at a spacing of 6 inches, so (6, 18) is the vertex. The height at 11 inches finds a.
The formula is
y = a(x − h)² + k
The vertex is (6, 18).
y(x) = a(x − 6)² + 18
Substitute (11, 68).
68 = a(11 − 6)² + 18
68 = a(5)² + 18
68 = 25a + 18
68 − 18 = 25a
50 = 25a
a = 5025
a = 2
y(x) = 2(x − 6)² + 18
Substitute x = 4.
y(4) = 2(4 − 6)² + 18
y(4) = 2(−2)² + 18
(−2)² = 4
2 · 4 = 8
y(4) = 8 + 18
y(4) = 26
Answer: y(x) = 2(x − 6)² + 18The plants are 26 inches tall.
Vertex form is a good fit. The greatest height is 36 cm, and it happens at a spacing of 10 cm, so (10, 36) is the vertex. The height at 13 cm finds a.
The formula is
y = a(x − h)² + k
The vertex is (10, 36).
y(x) = a(x − 10)² + 36
Substitute (13, 18).
18 = a(13 − 10)² + 36
18 = a(3)² + 36
18 = 9a + 36
18 − 36 = 9a
−18 = 9a
a = −189
a = −2
y(x) = −2(x − 10)² + 36
Substitute x = 8.
y(8) = −2(8 − 10)² + 36
y(8) = −2(−2)² + 36
(−2)² = 4
−2 · 4 = −8
y(8) = −8 + 36
y(8) = 28
Answer: y(x) = −2(x − 10)² + 36The plants are 28 cm tall.
Vertex form is a good fit. The least height is 10 cm, and it happens at a spacing of 14 cm, so (14, 10) is the vertex. The height at 18 cm finds a.
The formula is
y = a(x − h)² + k
The vertex is (14, 10).
y(x) = a(x − 14)² + 10
Substitute (18, 18).
18 = a(18 − 14)² + 10
18 = a(4)² + 10
18 = 16a + 10
18 − 10 = 16a
8 = 16a
a = 816
a = 12
y(x) = 12(x − 14)² + 10
Substitute x = 8.
y(8) = 12(8 − 14)² + 10
y(8) = 12(−6)² + 10
(−6)² = 36
12 · 36 = 18
y(8) = 18 + 10
y(8) = 28
Answer: y(x) = 12(x − 14)² + 10The plants are 28 cm tall.
Vertex form is a good fit. The greatest height is 24 feet, and it happens at 2 seconds, so (2, 24) is the vertex. The height at 3 seconds finds a.
The formula is
y = a(x − h)² + k
The vertex is (2, 24).
h(t) = a(t − 2)² + 24
Substitute (3, 20).
20 = a(3 − 2)² + 24
20 = a(1)² + 24
20 = a + 24
20 − 24 = a
−4 = a
a = −4
h(t) = −4(t − 2)² + 24
Substitute t = 0.
h(0) = −4(0 − 2)² + 24
h(0) = −4(−2)² + 24
(−2)² = 4
−4 · 4 = −16
h(0) = −16 + 24
h(0) = 8
Answer: h(t) = −4(t − 2)² + 24The ball was 8 feet high when it was released.
Vertex form is a good fit. The greatest profit is 480 dollars, and it happens at a price of 4 dollars, so (4, 480) is the vertex. The profit at 6 dollars finds a.
The formula is
y = a(x − h)² + k
The vertex is (4, 480).
P(x) = a(x − 4)² + 480
Substitute (6, 320).
320 = a(6 − 4)² + 480
320 = a(2)² + 480
320 = 4a + 480
320 − 480 = 4a
−160 = 4a
a = −1604
a = −40
P(x) = −40(x − 4)² + 480
Substitute x = 3.
P(3) = −40(3 − 4)² + 480
P(3) = −40(−1)² + 480
(−1)² = 1
−40 · 1 = −40
P(3) = −40 + 480
P(3) = 440
Answer: P(x) = −40(x − 4)² + 480The profit is 440 dollars.
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