Algebra: 6.1.1 Line of Best Fit
Solutions
Name __________________________________
Block ____ Date ___________
Bell Work: Classify as linear, quadratic or exponential, and then generate a regression equation for each data set.
Quadratic
Linear
Exponential
c.
a.
b.
x
1
3
5
7
9
11
13
y
5
7.2
8.9
11
13.3
14.8
17
𝒚 =. 𝟗𝟗𝟗 + 𝟒. 𝟎𝟎
x
-2
0
2
4
6
8
10
y
0.3
0.8
1.7
3.8
8.5
19.2
43.1
𝒚 = 𝟎. 𝟕𝟕(𝟏. 𝟓)𝒙
x
1
2
3
4
5
6
7
y
2.6
4
5.8
8.7
12.2
16.7
22
𝒚 = 𝟎. 𝟒𝟒𝒙𝟐 − 𝟎. 𝟎𝟎𝒙 + 𝟐. 𝟑𝟑
6-4. Sam collected data by measuring the pencils of her classmates. She recorded the length of the
painted part of each pencil and its weight. Her data is shown on the graph
at right.
a.
Describe the association between weight and length of the pencil. Remember
to describe the form, direction, strength, and outliers.
b.
Make a conjecture (a guess) about why Sam’s data had an outlier.
c.
Sam created a line of best fit: w = 1.4 + .25l where w is the weight of the
pencil in grams and l is the length of the paint on the pencil in centimeters.
(graph the line over the data) What does the slope represent in this context?
d.
Sam’s teacher has a pencil with 11.5 cm of paint. Predict the weight of the
teacher’s pencil.
e.
Interpret the meaning of the y-intercept in context.
6-5. Given the sequence 7, 11, 15, 19, …
a.
What kind of sequence is it?
b.
Define the sequence explicitly.
c.
Is 109 a term of the sequence? If so, which term is it?
6-6. Solve each system algebraically. Check by graphing or substitution.
8𝑦 − 𝑥 = 1
10𝑦 = 𝑥 + 5
a.
−𝑥 + 8𝑦 = 1
−𝑥 + 10𝑦 = 5
+
𝑥 − 8𝑦 = −1
−𝑥 + 10𝑦 = 5
(𝟏𝟏, 𝟐)
𝑦 = 2𝑥 + 10
(
5 2𝑥 + 10) = 2 − 6𝑥
10𝑥 + 50 = 2 − 6𝑥
16𝑥 = −48
𝒙 = −𝟑
8(2) − 15 = 1 
10(2) = 15 + 5 
8(2) − 𝑥 = 1
16 − 𝑥 = 1
−𝑥 = −15
𝒙 = 𝟏𝟏
2𝑦 = 4
𝒚=𝟐
5𝑦 = 2 − 6𝑥
𝑦 − 2𝑥 = 10
b.
6-7. Simplify each expression using only positive exponents.
3
a. (3xy )(–2y)
3 2
–2 –2
b. (x y )(x y )
c.
(−𝟑, 𝟒)
5(4) = 2 − 6(−3) 
4 − 2(−3) = 10 
𝑦 = 2(−3) + 10
𝑦 = −6 + 10
𝒚=𝟒
6𝑥 −2
–3
d. (–2x)
3𝑥 2
6-8. Solve each equation for the indicated variable.
a.
y = 3x + 2
(for x)
b.
𝑎=
𝑏
𝑐
(for b)
c.
6-9. Find the missing term in each sequence.
a.
b.
𝑥
2
−7 =
𝑦
3
(for x)
1
d. 𝑔 = 2 𝑎𝑡 2 (for t2)
y
1. Graph the function 𝑔(𝑥 ) = −(𝑥 − 3)2 + 4 without a calculator
8
6
4
2
x
−8
−6
−4
2
−2
4
8
6
−2
2. Find the equation of a line that contains (4, −7) and (0, 5).
−4
−6
Δ𝑦 −12
=
Δ𝑥
−4
−8
𝑦 = 𝟑𝒙 + 𝟓
𝑆𝑆𝑆𝑆𝑆 = 𝟑
3. Factor completely:
𝒙 – )
𝟑 𝒙 + )
𝟐 (
𝟐
3𝑥 2 −12 = (
3(𝑥 2 − 4)
4. Multiply the expressions using generic rectangles.
a. (3 − x)(1 + x)
b. (4x − 1)(x + 3)
+𝑥 3𝑥
−𝑥 2
+3 12𝑥
−3
3
−𝑥
4𝑥
−1
1
3
𝑥 4𝑥 2
−𝑥
−𝒙𝟐 + 𝟐𝟐 + 𝟑
−𝑥
𝟒𝟒𝟐 + 𝟏𝟏𝟏 − 𝟑
c. (3x − 5)2
−5 −15𝑥
3𝑥
9𝑥 2
3𝑥
25
−15𝑥
−5
𝟗𝟗𝟐 − 𝟑𝟑𝟑 + 𝟐𝟐
5. Solve the quadratic below twice: once by factoring and once by quadratic formula 𝑥 =
Verify that the solutions match.
−𝑏±√𝑏2 −4𝑎𝑎
2𝑎
.
x2 + 13x + 22 = 0
Quadratic Formula
Factoring
+2 2𝑥
𝑥 𝑥2
𝑥
22
11𝑥
+11
𝑥 2 + 13𝑥 + 22 = 0
(𝑥 + 2)(𝑥 + 11) = 0
𝑥 + 2 = 0,
𝒙 = −𝟐,
𝑥 + 11 = 0
𝒙 = −𝟏𝟏
𝑥=
−13 ± �132 − 4(1)(22)
2(1)
𝑥=
−13 ± √169 − 88
2
𝑥=
−13 ± √81
2
𝑥=
−13 ± 9
2
𝑥=
𝑥=
−13 + 9
= −𝟐
2
−13 − 9
= −𝟏𝟏
2
PARCC PREP
I.
𝒚 > −𝟐𝟐 + 𝟔
II
(𝑥 2 + 3)2 + 21 = 10(𝑥 2 + 3)
𝒖𝟐 + 𝟐𝟐 = 𝟏𝟏𝟏
−𝟏𝟏
𝟐𝟐
𝒖𝟐 − 𝟏𝟏𝒖 + 𝟐𝟐 = 𝟎
(𝑢 − 7)(𝑢 − 3) = 0
𝑢 = 7,
𝑥 2 + 3 = 7,
𝑥 2 = 4,
𝑥 = ±2,
𝟐
−𝟐
𝟎
𝑢=3
𝑥2 + 3 = 3
𝑥2 = 0
𝑥=0