Chapter 4 – section 1

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Transcript Chapter 4 – section 1

College Algebra

Acosta/Karwowski

Chapter 4

Quadratic equations and functions

Solving quadratic equations

CHAPTER 4 – SECTION 1

• • A quadratic is any equation with a 2 nd term and no higher degree terms degree In standard form ax 2 + bx + c = y • • Solving an equation means to isolate the x variable. Since there is an x 2 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 𝑥 2 ≠ 𝑥 for every x 2 𝑥 2 = |𝑥| thus: if 𝑥 2 = 25 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑥 = ±5

Reminders : working with square roots • • • If the square root is irrational – leave the radical form of the answer unless ASKED to round the answer - This is called the exact form of the number ex: 16 ans: 4 21 ans: 21 Simplify all radicals 21 𝑎𝑛𝑠: 21 80 𝑎𝑛𝑠: 4 5 Square root of a negative is an IMAGINARY number −4 = 2𝑖 not - no solution −12 = 2𝑖 3

Examples: solve

• • • • 5x 2 – 3 = 77 3x 2 + 7 = 25 (x – 4) 2 = 16 24 + (x – 3) 2 = 15 • x 2 +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 3x 2 + 5x B. You cannot do the square root on an expression that contains addition outside of () 𝑥 − 5 2 = 𝑥 − 5 𝑥 2 − 4 ≠ 𝑥 − 2

Using factors to solve quadratics

• • • Many quadratics will factor ex: x 2 – 7x + 12 = 3x 2 - 19x + 20 = 9x 2 – 121 = 15x 2 + 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 2. Factoring the quadratic • 3. Splitting the one equation into 2 and 4. Solving each equation

Examples

• • • • x 2 + 6x – 16 = 0 x 2 – 7x = -12 (x - 3)(x – 8) = 66 6x 2 + 29x = 5 • x 2 – 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 : x 2 + 2hx + h 2 = (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 3 rd term Fill in the blank x 2 + 8x + ___ x 2 – 10x + ___ x 2 + 5x + ____ Filling in this blank is called completing the square

• • Completing the square to solve equations 4x 2 – 40x - 20 = 0 Prepare for completing the square by dividing by 4 and moving the 20 ( which will be a 5) • • • • • x 2 – 10x = 5 Completing the square CHANGES the number so add the same amount to BOTH sides of the equation x 2 – 10x + 25 = 5 + 25 Factor and solve (x – 5) 2 = 30 x = 5 ± 30

Examples : solve the following

• 3x 2 – 24x + 6 = 0 • x 2 – 7x – 3 = 0 • 3x 2 – 11x – 4 = 0 • x 2 + 5x + 15

Quadratic inequalities in one variable

CHAPTER 4 – SECTION 3

What we already know

• • • 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

find f(x) = 0

x

find f(x)>0 find f(x)<0 So quadratic inequalities work like absolute value inequalities

• find x 2 – 9x >0

Examples

• Find 2x 2 + 7x – 30 < 0 • Find -2(x + 3) 2 + 2 > 0

• P 370(6-16)

Assignment

Guaranteed test question

CHAPTER 4 -

Guaranteed test question – worth 10 20 pts • • • Mathematics finds the formulas – You find a formula by solving without using any known numbers Solve ax 2 + bx + c =y for x thereby deriving a formula for x

• • • • • • Deriving the quadratic formula Solve ax 2 + bx + c = 0 by completing the square Divide by a :

x

2 

b a x a

0 Move constant to the right

x

2 

b a x

 

c a

Divide middle coefficient by 2 /square/ add

x

2 

b a x

b

2 4

a

2

a b

2 4

a

2 Get a common denominator and combine fraction Write as a square Isolate the x

x

2 

b a x

b

2 4

a

2 

b

2  4 4

a

2

ac x

b

2

a x

 2 

b

2  4 4

a

2

ac b

2  4

ac

2

a

Quadratic functions

CHAPTER 4 – SECTION 2

Quadratic functions

• • • f(x) = ax 2 + 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 – x 1 )(x – x 2 ) can be the same function also- called factored form

Standard form to vertex form

• Given f(x) = 2x 2 – 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) = x 2 – 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) = -2x 2 + 4x – 7 • method one – easiest – uses information from quadratic formula • 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) = -16x 2 + 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 Add to syllabus schedule -

Writing equations for quadratics

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

• • • • • • the quadratic function h(x) = ax 2 + 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/sec 2 ) an initial velocity force of 12 ft/sec and an initial location of 29 feet.

Banking formula- compound interest • • • • • P(r) = p 0 (1 + r) where p 0 t 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.

• 𝑥 = 9 • 2𝑥 − 7 = 3

Examples

3 𝑥 = 4 4 𝑥 = 5 • 10 − 2𝑥 = 𝑥 2 + 𝑥 • 𝑥 + 2 + 1 = 3𝑥 − 5

• P377(1-31)odd

assignment

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

• 𝑥 5 3 = −32

Examples

• 5𝑥 2 3 − 8 = 117

Quadratic like equations using 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 x 4 – 5x 2 = 6 𝑥 + 𝑥 = 12 Or If the equation is written as three “terms” with this same condition ex (x – 7) 6 + 2(x - 7) 3 - 15 = 0

Equations that become quadratic

Examples: solve

• 7

x

 6  9 5

x

 30  2 5 •

x

2  5 

x

7  2 

x

2 4

x

 7

x

 10 • 𝑥−3 𝑥+4 2𝑥+1 = 3𝑥+60

Assignment

• Supplement - p69 (1-9)(13-26)odd (30- 33)all