### Chapter 4 – section 1

```College Algebra
Acosta/Karwowski
Chapter 4
CHAPTER 4 – SECTION 1
• A quadratic is any equation with a 2nd degree
term and no higher degree terms
• In standard form ax2 + bx + c = y
• Solving an equation means to isolate the x
variable. Since there is an x2 term as well as
an x term this presents a challenge
algebraically.
•
Using roots to isolate the x
• If the x appears in the problem only once you can
isolate the x using inverses but first note
• In general the inverse of a power is its root
• But as seen in ch 5 square root is not the inverse
of square power unless we modify the functions
• you can cancel square powers with square roots
if you remember that
2
2 ≠  for every x
2
2 = ||
• thus: if  2 = 25   ℎ  = ±5
Reminders : working with square roots
• If the square root is irrational – leave the radical form of
called the exact form of the number
ex: 16 ans: 4
21 ans: 21
21 : 21
80 : 4 5
• Square root of a negative is an IMAGINARY number
−4 = 2 not - no solution
−12 = 2 3
Examples: solve
• 5x2 – 3 = 77
• 3x2 + 7 = 25
• (x – 4)2 = 16
• 24 + (x – 3)2 = 15
• x2 +8x = 15
Draw back to inversing
• If x does not appear in the problem only once
using square root to isolate the x becomes
problematic since:
• A. You cannot combine unlike terms
3x2 + 5x
• B. You cannot do the square root on an
expression that contains addition outside of ()
−5 2 = −5
2 − 4 ≠  − 2
• ex: x2 – 7x + 12 =
3x2 - 19x + 20 =
9x2 – 121 =
15x2 + 35x – 90 =
• Mathematical fact if ab = 0 either
a= 0 or b = 0
Thus: zero product rule is born
• Given a quadratic equation, you can
SOMETIMES isolate the x by
1. Making the equation = 0
3. Splitting the one equation into 2
and
4. Solving each equation
•
Examples
•
•
•
•
x2 + 6x – 16 = 0
x2 – 7x = -12
(x - 3)(x – 8) = 66
6x2 + 29x = 5
• x2 – 10 = 0
Bridging the gap
• Some problems cannot be done by factoring
because they do not have integral factors.
• These problems must be re written so that the
x only appears in the equation once.
• The process is called completing the square
Completing the square
• Concept - write a quadratic using the variable x only
once • tools :
x2 + 2hx + h2 = (x – h)2
• so - a quadratic of this form can be written with only
one x
• Given the first 2 terms you can determine the 3rd term
•
Fill in the blank
x2 + 8x + ___
x2 – 10x + ___
x2 + 5x + ____
Filling in this blank is called completing the square
Completing the square to solve
equations
• 4x2 – 40x - 20 = 0
• Prepare for completing the square by dividing by 4 and
moving the 20 ( which will be a 5)
• x2 – 10x
= 5
• Completing the square CHANGES the number so add
the same amount to BOTH sides of the equation
• x2 – 10x + 25 = 5 + 25
• Factor and solve
•
(x – 5)2 = 30
x = 5 ± 30
Examples : solve the following
• 3x2 – 24x + 6 = 0
• x2 – 7x – 3 = 0
• 3x2 – 11x – 4 = 0
• x2 + 5x + 15
CHAPTER 4 – SECTION 3
• The solutions of inequalities in one variable
are intervals on the number line
• The solutions to inequalities are directly
related to the solution of equations
• There are 2 solutions to quadratic equations.
What the graph of the quadratic
function tells us
y
•
x
find f(x) = 0
find f(x)>0
find f(x)<0
So quadratic inequalities work like absolute
value inequalities
Examples
• find x2 – 9x >0
• Find 2x2 + 7x – 30 < 0
• Find -2(x + 3)2 + 2 > 0
Assignment
• P 370(6-16)
Guaranteed test question
CHAPTER 4 -
Guaranteed test question – worth 1020 pts
• Mathematics finds the formulas –
• You find a formula by solving without using
any known numbers
• Solve ax2 + bx + c =y
for x thereby
deriving a formula for x
Solve ax2 + bx + c = 0 by completing the square
• Divide by a :
• Move constant to the
right
• Divide middle coefficient
• Get a common
denominator and
combine fraction
• Write as a square
• Isolate the x
x 
2
b
c
x
a
x 
2
b
a
x
a
x 
2
x 
2
b
0
c
a
b
x
2
a
4a
b
b
x
a
b

a
2
4a
2

2
c
4a
b  4 ac
2

2
4a
2
b 
b  4 ac

x




2
2a 
4a

x
b 
2
b  4 ac
2
2a
2
2
CHAPTER 4 – SECTION 2
• f(x) = ax2 + bx + c = y is a quadratic function
called standard form
• f(x) = a(x – h)2 + k = y can be the SAME
quadratic function - called vertex form
• f(x) = (x – x1)(x – x2) can be the same function
also- called factored form
Standard form to vertex form
• Given f(x) = 2x2 – 12x + 10
Related topic for circles
• Standard form of circle (transformation form)
• Get standard form from simplified form.
• homework p 19(30-35)
Function concepts- what the equation
and graph tell us
•
•
•
•
•
•
x and y intercepts
f(x) = 0 is the x- intercept
f(0) = y is the y - intercept
maximum/minimum of a parabola
called vertex
related to line of symmetry
(a vertical line through the vertex)
intervals of increase or
decrease
f(#) = y (evaluate for y)
f(x) = # (solve for x)
Concavity – (also related to max/min)
Sketching a graph from the equation
Concavity
• Oriented up/ down
• Width
Examples: find x and y intercepts
• f(x) = x2 – 12x +32
also find f(3) and f(x) = 12
• g(x) = (x + 2)(x – 9) also find g(7) and g(x) = 2
• k(x) = 3(x – 6)2 - 15 also find k(4) and k(x) = 9
Find vertex(ie: max/min) / line of
symmetry/ intervals of inc. or dec.
•
g(x) = 3(x + 5)2 + 3
Finding vertex etc.
• f(x) = -2x2 + 4x – 7
• method one – easiest – uses information
• Method two – change to vertex form -
Example: word problem
• A ball is thrown into the air has a height given
by the function h(x) = -16x2 + 34x + 15
•
•
•
•
•
Find the y intercept and interpret its meaning?
When will the ball hit the ground?
Where will the ball be in 1.2 seconds?
When will the ball be 17 feet above ground?
How high did the ball go?
Sketching a graph for a quadratic
Assignment
•
•
•
•
sec 4.1 P326(1-33)(56 – 61) all
P 354 (5-38) odd add 355(46-50)
Worksheet
CHAPTER 4 - SUPPLEMENT
Writing equations – word problems
• To write an equation for a quadratic you need either
• 3 random points
•
or the vertex and one other solution point
• Or factors and one other point
• or a formula which has already been derived by
someone else.
Physics formula
•
h(x) = ax2 + bx + c
• has a physics application where
•
a = acceleration of an object after it is released (usually
this is gravity)
•
b = its initial velocity (force with which it is released)
•
c = the original location of the object
• Write a function for an object with gravitational
acceleration (known to be -16 ft/sec2) an initial velocity
force of 12 ft/sec and an initial location of 29 feet.
Banking formula- compound interest
• P(r) = p0 (1 + r)t
• where p0 is original amount in bank
•
t is a set amount of time
•
r is the rate of interest
•
note: t must be the same units as r
if r is 6% per month then t = 7 is 7 months
Ex. Mark invests \$4500 for 3 years. Write a
function for his account as a function of r.
Given 3 point
• (2,13) (-3, 38) ( 1,6)
•
Given vertex and one point
• (2,4) is the vertex and the graph goes through
the point (3,10)
Given factors/x intercepts
• x – intercepts can be found from factors so
factors can be found from x- intercepts - even
if those intercepts are irrational
• Factors can also be found given solutions that
are not intercepts – ie solutions that are
imaginary – note that imaginary and irrational
solutions come in pairs (conjuagates)
• Note that a third point is necessary to find the
stretch factor
examples
• Given the points (5,0) (-2,0) and the point (3,-2) write
an equation – you could do a system of equations since
this is 3 points
• Or •
a(x – 5) (x + 2)= 0 must be true
• Thus a(x -5)(x +2) = y is the basis for the equation
and
•
a(3 – 5)(3 +2) = -2
so -10a = -2 and a = 1/5
answer g(x) = 1/5 (x – 5)(x + 2)
Examples
•
•
•
•
given intercept 2 + 3, 0 and point (4, -3)
You know there is another intercept –
You have factors
You can simplify the quadratic and find the
stretch factor
example
•
•
•
•
Given solution - x = 3i and point (3, - 2)
You know another solution is x = - 3i
You have factors
You simplify and find the stretch factor
assignment
• P328 (62,63,66,67)
• Worksheet – writing equations for quadratics.
Solving root equations
CHAPTER 4 – SECTION 4
The inverse of a root is it’s
corresponding power
• Algebraic solving involves inversing an
operator or function to determine the input
value.
•
Examples
3
•
=9
=4
•
2 − 7 = 3
•
10 − 2 =  2 +
•
+2+1=
3 − 5
4
=5
assignment
• P377(1-31)odd
Other equations – quadratic in nature
SUPPLEMENT F
Powers/roots generalized
• Rational exponents
• The inverse of x a/b is x b/a
• Quadratic like: a trinomial in which the degree
of the leading term is twice the degree of the
middle term can be solved like a quadratic
Examples
5
3
•  = −32
2
3
• 5 − 8 = 117
substitution
• If the equation has 3 terms where the middle
exponent is half of the leading exponent then
it is like a quadratic equation and can be
solved using a temporary substitution
ex
x4 – 5x2 = 6
2 3 +  1 3 = 12
• Or If the equation is written as three “terms”
with this same condition
ex (x – 7)6 + 2(x - 7)3 - 15 = 0
Examples: solve
•
•
•
7
x6
2
x5
−3
+4


=
9

5 x  30
7
x2

2
5
4x
x  7 x  10
2+1
3+60
2
Assignment
• Supplement - p69 (1-9)(13-26)odd (30- 33)all
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