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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Algebra 1
Unit 7: Post Keystone Exam Topics
Lesson 1 (PH Text 4.2, 5.8, 9.1, 7.6, 10.5): Families of Functions
Lesson 2 (PH Text 9.3): Solving Quadratic Equations
Lesson 3 (PH Text 9.4): Solving Quadratic Equations by Factoring
Lesson 4 (PH Text 5.8): Solving Absolute Value Equations
Lesson 5 (PH Text 2.5): Solving Literal Equations
Lesson 6 (PH Text 10.4): Solving Radical Equations
Lesson 7 (PH Text p.605): Midpoint and Distance Formulas
Lesson 8 (6.0 only) (PH Text 10.2-3): Operations with radicals involving variables
Lesson 9 (6.0 only) (PH Text 10.2-3): Division of Radicals involving conjugates
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 1 (PH Text 4.2, 5.8, 9.1, 7.6, 10.5): Families of Functions
Objective:
to identify families of functions for equations and graphs
to predict what the graph of an equation will look like
Explore - Using a Graphing Calculator:
1) Graph each function. Sketch a picture of each graph.
y  x2  6
y  x2
y  x 4
y  7x
y  x2
y  2x
y   4x 1
y  6 x
y  x 3
y  x2  1
y  3x 2
y  3 x
y4 3
2x
y
1
x2
2
y   x5
2
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Summarize:
2) Sort the graphs into 5 Categories by grouping how they look.
3) What similarities among the graphs do you see in each category?
4) What differences do you see?
These five Families of Functions are:
Type
Linear
Quadratic
Exponential
Graph
Equation
Special
Characteristics
3
Absolute
Value
Radical
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Families of Functions:
Linear function highest power of x is _________
graph forms a __________________
positive coefficient of x slants _________
negative coefficient of x slants _________
Absolute Value function has an absolute value symbol around a variable expression
graph forms a _________
positive coefficient of x opens _________
negative coefficient of x opens _________
Quadratic function highest power of x is _________
graph forms a __________________, or a _________
positive coefficient of x opens _________
negative coefficient of x opens _________
Exponential function In the form y = a · bx, where a ≠ 0, b > 0, b ≠ 1,
and x is a real number
graph forms a __________________
positive coefficient of x opens _________
negative coefficient of x opens _________
Radical function highest power of x is _________
graph forms a _____________________________
positive coefficient of x opens _________
negative coefficient of x opens _________
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Class Practice:
Sketch an example of each of the following:
1.) Quadratic
2) Linear
3) Radical
Identify the Function Family to which each belongs. Describe how you knew:
4) y  5 x  4
5) y  7 45 x
6) f ( x)  5x  7
7) y = 6x2 + 1
8) y = 4x – 1
9) y = x2 + 3x + 2
10) y = 3x
11) y  8x
12) y = 7 – x
5
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 2 (PH Text 9.3): Solving Quadratic Equations
Objectives:
to solve quadratic equations by graphing and using square roots
A quadratic equation is any equation that can be written in the form ax2 + bx + c = 0, where a ≠ 0.
The standard form of a quadratic equation is ax2 + bx + c = 0.
Quadratic equations can be solved in a variety of ways. We will consider graphing, using square roots, and
factoring.
Solving quadratic equations by graphing:
Graph the function. The x-values of points where the graph crosses the x-axis are considered the solutions.
Solving quadratic equations by using square root:
Isolate the squared term. Find the square roots of each side, and simplify.
2x2 – 98 = 0
HW: p.551 #20-36 even
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 3 (PH Text 9.4): Solving Quadratic Equations by Factoring
Objectives: to solve quadratic equations by graphing and using square roots
Finding solutions to quadratic equations is also an important use of factoring. A replacement set is the set of
all solutions to a polynomial equation. Any real numbers that make an equation a true statement are a part of
the equation’s replacement set.
The easiest way to find an equation’s solution set is to apply the zero-product property.
Zero-Product Property:
For all real numbers a and b, if ab = 0 then a = 0, b = 0 or both a and b = 0.
To find the replacement set to a given equation:
1. Write the equation in standard form.
2. Factor the polynomial completely.
3. Set each of the factors equal to zero.
4. Solve for the variable.
5. Check by replacing the potential solution for the variable in the equation. If it results in a true
statement, it is a part of the equation’s replacement set.
Example:
8x2 + 10x – 3 factored is (2x + 3) (4x – 1)
so (2x + 3) = 0
Check:
Replacement set for 8x2 + 10x – 3:
Find the possible solutions for 4x2 – 21x = 18
HW: p.558 #8-36 even
7
and/or
(4x – 1) = 0
{
}
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
9
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 4 (PH Text 3.7): Solving Absolute Value Equations
Objective:
to solve equations that involve absolute value
Some equations may have two solutions. One time this can happen is when the variable is within absolute
value bars.
two solutions
one solution
no solution
x5
x  5
Examples:
x 5
x = 5 or -5
x=5
absolute value can
never be negative
Class Practice:
x  16
1)


2)
x  5  11
5)
6 
3)
3 w 4
6)
x  5  9

4)
4 n  32
7)
x9  3
9)
x
2
8)
p3 4  0
Jeff estimates his stride is 16 inches. However, any given stride is likely to vary from this estimate by up to
2 inches. Write and solve an equation to find Jeff’s minimum and maximum stride length.
HW: p.211 #10-30 even
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 5 (PH Text 2.5): Solving Literal Equations
Objective:
to use the strategies of reciprocals and opposites to solve equations involving variables only
A literal equation is an equation that expresses a relationship among variables.
Formula - shows the relationship between two or more variables
You can transform (change) a formula to define a different variable by “solving for” that variable. Use the
skills we learned to solve equations.
Example:
I = prt
I
p
rt
Class Practice: Solve for the underlined variable. Show your work!

1) x  10  y
2) 3(2a  b)  c
3) h  2(l  2m)
5)
a  2b
 4b
3c
6) x  y  5 
x
2
Solve each for the given variable.
7)
P  2l  2w
for w
8)
1
A  bh
2

11
for b
9)
d = rt
for t
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 6 (PH Text 10.4): Solving Radical Equations
Objective:
to solve equations containing radicals;
to identify extraneous solutions.
For any real number n,
n2  n
Remember:
Simplify.
a)
90
b)
4
18
c)
3( 6  2 8)
b)
27x3
c)
50a6
Example: Simplify.
a)
72 y 2
Class Practice:
Simplify.
1)
8b8
2)
63x3
3)
5c5
4)
300x6
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Example 2: Evaluate 2n  10 a) for n = 3
b) for n= 9
Class Practice:
Evaluate for the given variable. Then simplify, if possible.
5)
c  7; c  15
Example 3: For what values of y will
6)
5 x  6; x  6
2 y  5 be a real number?
Class Practice:
Find all values of x that make each radical expression a real number.
7)
5x  10
8)
9)
4 x  2
10)
14
2x  7
3x  9
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
My Class Practice:
Find all values of x that make each radical expression a real number.
1)
2)
x  3  11
4x  7  1
3)
6x  4  4x  6
4)
x  4 x  12
5)
12  6 x
6)
3x  13  7 x  3
7)
2 x  3  13
8)
3x  8  x  6
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 7 (PH Text p.605): Midpoint and Distance Formulas
Objectives:
To find the distance between two points in a coordinate plane;
To find the midpoint of two points.
1. Graph the points ( –3, 4 ), ( 1, 1 ), ( –3, 1) and connect them
to form a triangle. Mark the lengths of the legs by counting units.
Use the Pythagorean Theorem to
find the length of the hypotenuse.
a 2  b2  c2
Now use the distance formula to find
the length between (1,1) and ( –3, 4 ).
d=
 x2  x1 
2

 y 2  y1

2
The distance formula: For points P  x1 , y1  and Q  x2 , y2  in the coordinate plane, the distance d between
the points is given by:
d=
 x2  x1 
2

 y 2  y1

2
Round answers to the nearest tenth!!!!
2. Find the distance between (1, 4) and (−2, −5).
3. Find the distance between (−3, 2) and (3, −2).
4. One hiker is 4 miles west and 3 miles north of the campground. Another is 6 miles east and 3 mile
south of the campground. How far apart are the hikers? (the camp ground is at (0, 0) )
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
5. Mickey travels 15 miles west, then 20 miles north. Jamie travels
5 miles east, then 10 miles south. How far apart are they?
6. Quadrilateral KLMN has vertices with coordinates K(-3, -2), L(-5, 6), M(2, 6) and N(4, -2).
a. Show that LK  MN .
b. Use slopes to show that LK and MN are parallel.
The midpoint of a segment is the halfway point between two endpoints. The coordinates of a midpoint are
the averages of the coordinates of the endpoints.
The midpoint formula: For endpoints P  x1 , y1  and Q  x2 , y2  on the coordinate plane the midpoint m can
be expressed by:
x x y y 
M = 1 2 , 1 2 
2 
 2
8. Find the midpoint of P(−1, 6) and Q(5, 0)
7. Find the midpoint of A(2, -1) and B(4, -3)
HW: p.605 #1-6
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
 x2  x1    y2  y1 
The distance formula:
d
The midpoint formula:
 x  x y  y2 
M 1 2, 1

2 
 2
2
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2
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 8 (6.0 only) (PH Text 10.2-3): Operations with radicals involving variables
Objective: to add, subtract, multiply and divide radicals involving variables.
To add and subtract, simplify first, then combine like terms.
Example: Simplify. Make sure all answers are in simplest radical form.
1) 100 x 2 y  144 x 2 y  5 x 2 y
2) 18 x 2 y  8 x 2 y  50 x 2 y
Remember when multiplying, multiply the coefficients (number outside) then multiply the radicands (the
number inside). Simplify your answers if possible.
Example: Simplify. Make sure all answers are in simplest radical form.
a a
3)  7 y 3 8 y
4) 5a
b b
To divide, simplify first using power rules, then rationalize the denominator if necessary.
Example: Simplify. Make sure all answers are in simplest radical form.
5)
1
x7
6)
20
24 x 2 y
3 xy 2
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Class Practice: Perform the indicated operations. Express your answers in simplest radical form.
2) 175a2b  3 112a2b
72 xy 2  2 98 xy 2
1)

3) 3 81ab2  2 ab2  5 9ab2
4) 3 n
5) 2 8xy
3xy
6)

8)
7)
9)

x 2 y
2
27 x 5
3x
10)
21

2
a5b bc2
3
a3
cd
c3d 3
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Lesson 9 (6.0 only) (PH Text 10.2-3): Division of Radicals involving conjugates
Objective: Be able to rationalize denominators using conjugates.
The radical expression 3  2 2 is called the conjugate of 3  2 2 . You use this to rationalize the
denominator of a fraction when adding and subtracting is involved.
If m and n are non-negative, then the binomials a m  b 2 and a m  b 2 are conjugates.

The sum and difference of the same two terms.
Class Practice:
Write the conjugate of each binomial.
1) 1  2
3) 5  3 5
2 ) 2 3
Example: Rationalize the denominator.
5
a)
3 2
Class Practice:
Rationalize the denominator and simplify.
2
5 1
4)
5)
6 3
5 3
7)
5
2 11  2
8)
7
3 10  5
25
b)
2 5 3
3 2 2
6)
2 3
1 5
9)
3 22
3 2 5
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Section 4: Solving Radical Equations.
Objective: Be able to solve radical equations
A radical equation is an equation that contains radicals with variables in the radicand.
To solve:
1)
2)
3)
4)
Isolate the radical on one side of the equation and combine any like terms
Square both sides to eliminate the radical
Repeat steps 1 and 2 if necessary.
Check your answer.
Example: Solve and Check.
1) 3  a  1  5
2)
3) 3 3x  2  5 x
26
5t 2 16  t
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
Class Practice: Solve and check.
1)
a4 6
3) 6  3 y  0
2)
4x  4  0
4)
5x  1  3  7
5)
3x  4  1  4
6)
3n2  12  3n
7)
2m
5  7
3
8)
5 y2  7  2 y
27
Algebra 1
Mrs. Bondi
Unit 7 Notes: Post Keystone Exam Topics
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