Unit 4: Radicals and Rationals
Transcript of Unit 4: Radicals and Rationals
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Unit 4: Radicals and Rationals
Shreya Nadendla, Sahana Devaraj, Divya Hebbar
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¿What is a RATIONAL function?
A rational function is any function which can be defined by a rational fraction. It is an algebraic fraction such that both the numerator and the denominator are polynomials. The coefficients of the polynomials do not need to be rational numbers.
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¿What is a RADICAL function?
A radical is something that is under a radical sign, i.e. a square root or cube root. A radical function contains a radical expression in the numerator with the independent variable (usually x) in the radicand. Usually radical equations where the radical is a square root is called square root functions.
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Direct Variation
Definition- A direct variation is a relationship in which the ratio of y to x is a constant, k. We say y varies directly with x.
Formula- Y= kx , k≠0 x and y →variables ,
K → constant of variation
EX: Y= 3x
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Direct Variation -Real World Example
Brady bought an energy efficient washing machine for her new apartment. If she saves about 10 gallons of water per load, how many gallons of water will she save if she washes 20 loads of laundry?
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Inverse VariationDefinition- An inverse variation is a relationship in which the product of two variables is equal to a constant. We say y varies inversely with x.
Formula- Y= k/x , k≠0 x and y →variables ,
K → constant of variation
EX: Y= 250/x
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Inverse Variation- Real World Example
If 16 women working 7 hours a day can paint a mural in 48 days, how many days will it take 14 women working 12 hours a day to paint the same mural?
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The three different values are inversely proportional; for example, the more women you have, the less days it takes to paint the mural, and the more hours in a day the women paint, the less days they need to complete the mural.
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Joint VariationDefinition- A joint variation is a relationship in which y varies directly with the product of x and z. We say y varies jointly with x and z.
Formula- Y= kxz , k≠0 x and y →variables ,
K → constant of variation
EX: Y= 5xz
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Joint Variation- Real world Example
The area of a triangle is jointly related to the height and the base. The dorito company is making a new flavour of doritos, called drool ranch. The base of the new chip will be increased by 40% and its height decreased by 10%, what will be the percentage change in the area of the new dorito?
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We know that the equation for the area of a triangle is A=12bh, (b = base and h = height) so we can think of the area having a joint variation with b and h, with k=½.
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¿What is a RATIONAL expression?
A rational expression is the ratio of two polynomials. It is a fraction whose numerator and denominator are polynomial expressions. Any expression made out of the 4 "rational operations" +, -, /, and * is considered to be rational. An improper rational expression is where the degree (highest power of x) of the numerator is greater than the degree of the denominator.
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Adding Rational ExpressionsAdding rational expressions is similar to adding fractions. You need a common denominator between the two expressions to do so.
1. Create similar denominators 2. Subtract the numerators, and write the result over the common denominator
Ex: Add: 1/x + 1/y.In this example, the LCD=xy. To obtain equivalent terms with this common denominator, multiply the first term by y/y and the second term by x/x.
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Subtracting Rational Expressions Subtracting rational expressions is similar to subtracting fractions. You need a common denominator between the two expressions to do so.
1. Create similar denominators 2. Subtract the numerators, and write the result over the common denominator
Ex: Subtract: 1/y− 1/(y−3).Since the LCD=y(y−3), multiply the first term by 1 in the form of (y−3)(y−3) and the second term by yy.
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Step 1: Factor all denominators to determine the LCD.
Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by (x−5)(x−5) and the second term by (x+3)(x+3).
Subtracting rational expressions Ex. 2
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Step 3: Add or subtract the numerators and place the result over the common denominator.
Step 4: Simplify the resulting algebraic fraction.
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Multiplying Rational Expressions● When multiplying fractions, we can multiply the
numerators and denominators together and then reduce.
Solution: Factor the denominator x2−25 as a difference of squares. Then multiply and cancel.
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Dividing rational Expressions
Solution: First, multiply by the reciprocal of the divisor and then cancel.
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Solution: The trinomial −2x2+x+3 in the numerator has a negative leading coefficient. Recall that it is a best practice to first factor out a −1 and then factor the resulting trinomial.
Multiplying Rational Expressions Ex. 2
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Solution: Begin by multiplying by the reciprocal of the divisor. After doing so, factor and cancel.
Dividing Rational Expressions Ex. 2
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Solution: Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, x2−36.
Simplifying Rational Expressions
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Simplifying Rational Expressions EX.2Step 1: Begin by factoring the denominator
Step 2: We can see that the LCD is (y+1)(y−1). Find equivalent fractions with this denominator.
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Step 3: Subtract and add the numerators and place the result over the common denominator.
Step 4: Finish by simplifying the resulting rational expression.
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What is a COMPLEX fraction?
● To simplify, divide the top fraction by the bottom
(multiply top by reciprocal of bottom)
¾ ÷ ½= ¾ * 2/1 =6/4 = 1 ½
● fraction where the numerator or denominator consists of one or more fractions.
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Method 1: Simplifying Using Division
Solution: The LCD of the rational expressions in both the numerator and denominator is x2. Multiply by the appropriate factors to obtain equivalent terms with this as the denominator and then subtract.
● The division method involves multiplying by the reciprocal of the divisor.
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Method 2: Simplifying using the LCD ● The LCD method involves multiplying both the numerator
and denominator by the LCD of all the given fractions.
Solution: Considering all of the denominators, we find that the LCD is x2. Therefore, multiply the numerator and denominator by x2:
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It takes 8 days for Sahana to finish a math project and it takes 2 days for Shreya and 4 days for Divya to finish a Math Project. How long would it take to finish a Math Project if Sahana, Divya and Shreya worked together?
Real World Problems - Working Together
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Solution
x/4 + x/2 + x/8 = 1 8( x/4 + x/2 + x/8 =1 ) 2x + 4x + 1x = 8 7x=8 x= 8/7 WE can complete the math project in 1 and 1/7 days if they work together.
Work Rate Time Work done
Sahana 1/8 x x/8
Shreya 1/2 x x/2
Divya 1/4 x x/4
Fraction of job done when working alone in one hour
We all work for the same amount of time to complete the project
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A motorboat makes an upstream trip on a river in 3 hours against the current, which is of 2 miles per hour.The return downstream trip with the same current takes 2 hours. Find the motorboat speed in still water and the trip length.
-Let ‘u’ denote the motorboat speed in still water in miles per hour.
Real World Problems - Wind Speed Rational
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Solution
Let u denote the motorboat speed in still water in miles per hour.
Then the speed of the motorboat is u - 2 relative to the river's banks when it moves upstream, and u + 2 when it moves downstream.
The length of the upstream trip is equal to 3*(u - 2) miles because 3 hours to go up
The length of the downstream trip is equal to 2*(u + 2) miles because 2 hours to go down
Since it is the same length, this gives you an equation with one unknown
3*(u - 2) = 2*(u + 2).
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Solution(Continued)
Let's open parentheses, collect variable terms on the left side, constant terms on the right side and reduce like terms, step by step:
3u - 6 = 2u + 4,
3u - 2u = 6 + 4,
u = 10.
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Work Rate - Practice
It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it take John and Linda, work together, to mow the lawn?
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Solution
● We first calculate the rate of work of John and Linda ● John: 1 / 1.5 and Linda 1 / 2 ● Let t be the time for John and Linda to mow the Lawn. The work done by John alone is given by ● t * (1 / 1.5)● The work done by Linda alone is given by ● t * (1 / 2)● When the two work together, their work will be added. Hence ● t * (1 / 1.5) + t * (1 / 2) = 1● Multiply all terms by 6 ● 6 (t * (1 / 1.5) + t * (1 / 2) ) = 6● and simplify ● 4 t + 3 t = 6● Solve for t ● t = 6 / 7 hours = 51.5 minutes.
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Work Rate - Upstream and Downstream1. Maxim is kayaking in the Russian
River which flows downstream at a rate of 1 mile per hour. He paddles 5 miles downstream and then turns around and paddles 6 miles upstream. The trip takes 3 hours. How fast can Maxim paddle in still water? Hint: there will be 2 answers and 1 is negative and it is impossible to paddle at a negative speed, which makes this an extraneous solution. So the other solution is correct.
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Solution
After multiplying by the LCD, distributing the LCD and reducing you should get:
3x^2 - 11x - 4=0 x=4 and x=-⅓
It would be impossible to kayak at a speed of -⅓ mph so the other answer which is 4 would be the answer. -⅓ would be an extraneous solution and the answer is 4 mph.
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Parts of a Radical Equation when Graphing-
y=a√ b(x-h)+k
Vertical Stretch/ horizontal shifts right/ shifts graphVertical Shrink stretch/ left up/down Shrink
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Transformations
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How to Solve Radical Equations
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Parts of A Rational Equation When Graphing
y= -a/(x-h)+ k
Shifts graph up/down
Shifts graph left when positive, right when negative
reflects
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Transformation
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y=-1/(x-4)+7 List Transformations
1. Reflection over x-axis
2.4 units right
3.7 units up
Transformation
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Discontinuities
Discontinuities are holes and vertical asymptotes because they create a break in the graph.
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Holes
Holes are produced when the numerator and denominator share a factor
1. To find the x-coordinate, set the factor that is common to both the numerator and denominator of the function equal to 0. Solve to find the x coordinate. x+2=0, x= -2
2. simplify the function by getting rid of the common factor. Since we are simplifying it, it becomes a “new” function since the domains will change.
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Holes (cont.)
3. to find the y-coordinate of the hole, replace x in the simplified function with the x-coordinate of the hole (-2).
4. The y coordinate of the hole is -¼, so the coordinates of the hole are (-2, -¼)
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Vertical Asymptotes
Vertical Asymptotes
Vertical Asymptotes
Vertical Asymptotes
To find the VA, get rid of any common factors, set the denominator=0 and solve for x.
x-2=0, VA is at x=2
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Vertical Asymptote Practice
1. (x-3)/(x-3)(x-4)2.
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Domain
To find the domain, find the vertical asymptote and any holes first. We know that this function has a hole at (-2, -.25) and a VA at 2.
TO
So, the domain will break at -2, and 2. Using interval notation, the domain= (-∞,-2)∪(-2, 2)∪ (2,∞)
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Range
Just like domain, except, exclude the y-value of the hole and the horizontal asymptote.
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X and Y intercepts
To find the y-intercept (the point where the graph crosses the y-axis), substitute in 0 for x and solve for y or f(x).
To find the x-intercept(s) (the point where the graph crosses the x-axis or zeros), substitute in 0 for y and solve for x.
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Horizontal Asymptotes
There are 3 different cases
If the degree of the numerator is more than the degree of the denominator, there is no Horizontal Asymptote
If the degree of the numerator is less than the degree of the denominator, then the Horizontal Asymptote= 0
If the degree of the numerator = the degree of the denominator, then you divide the coefficients to get your asymptote.
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Horizontal Asymptote Practice
1. x^2+2x+8/x2. x^3+3x+5/x^4+6x3. 6x^2+8/2x^2+10
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LimitLimit is all about what happens to x as it becomes very big in the positive or negative direction.
The first thing to do when finding limit, it figuring out the Horizontal Asymptote
So the horizontal asymptote would be 2. This means that as x becomes very, very large in the positive or negative direction, it will become closer and closer to the line y=2
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Limit (cont.)
Graph the function so you know what it does as it gets bigger and smaller
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Limit practice
1. What is the limit for 2x^2/ 4x^2 + 3x 2. What is the limit for 2x/ x^2 + 4x
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