4 markers 2 erasers One box per group ( star group of 6) 1 pencil (just in case) Some small post-it notes 1 glue stick One person from each group collect all items and place them back into the box.
Concept Category 1 Analyzing Functions From a graph From a table From an equation
Part 1 - Key Features: Domain & Range End-behavior Is the graph even or odd? Where on a graph is the function increasing or decreasing? Where on a graph is the function positive or negative?
Pair Share: What do you remember?
.
Or, how big or small can x-values be?
Pair Work: Find the Domain, Range, End-Behavior #1 #2
Pair Share 2: What do you remember? Where on the graph is the function increasing or decreasing? Where on the graph the function is positive or negative?
Interval of increase: x Interval of decrease: Positive: Negative: Did you get? 2, 0.8 x (, 2) U (0.8, ) 2 x 0.8 ( 2, 0.8) x 3 x 1, 2 ( 3, 1) U (2, ) x 3, 1 x 2 (, 3) U( 1, 2)
Pair Work 2 x y Find the intervals of increase and decrease. Find (estimate) where the function is positive or negative. x
Pair Work 3:
8/30 th Warm-Up (Algebra 2 Review) f(x) is the relation shown below. Decide if the following statements are True or False and explain why a.) This is a function b.) f(-2) = f(4) c.) 3 < d.) -2 = f(2) e.) 5 > f(7) f.) If f(x) = 1 then x = -3
SOLUTION: a.) This is a function (T) b/c each x-value is linked to exactly one y-value b.) f(-2) = f(4) (F) b/c f(-2)= 1 and f(4)= -1 and 1-1 c.) 3 < y-value of the y-intercept (F) b/c y intercept is (-2,-2) y-value of the y-intercept is -2; and 3 is not < -2 d.) -2 = f(2) (F) b/c f(2) = -3 and -2-3 e.) 5 > f(7) (T) b/c f(7) = -3 and 5>-3 f.) If f(x) = 1 then x = -3 (F) b/c f(-3) = 5 not 1
Finding domain and range from a table:
Did you get? the domain for this table of values would be {1, 2, 3, 4} and the range would be {6, 11, 16, 21}.
Pair Share 3: What can you tell each other about the key features of function f ( x) x 3? 2 Hint: some people find that a sketch helps.
Finding key features given a function: Remember Transformations? Or, you can use a table chart Domain : (, ) Range : (,3] End Behavior : left x, y right x, y Interval of increase : (,0) Interval of decrease : (0, ) Positive at : ( 1.8,1.8) Negative at : (, 1.8) U(1.8, )
Pair Work 4: Find all the key features of this equation f ( x) ( x 3) 4 2
Solution Domain: Range: End-Behavior: Interval of increase: Interval of decrease: Positive at: Negative at:
Rules of Transformation 2
Updates? Please write down what you remember from last Friday s lesson: Provide an example
Concept Category 1 Part 2 Transformations Graphically Algebraically
Warm-up Given the graph of f(x) below, graph -3 f( x 2) 1. y x
One of the methods. Table-Chart
Happy Tuesday Cell-phone in your bag Notebook Pens/Pencils Learning Targets Markers White board Transformations II Rational Graph Review Assignment: pg308 #13-22
Alg 2 Review: Transformation Pair work Sketch the equations below What do you remember about parent function graphs from last year?
2( x) 2 Solution: Right 3 units Down 4 units
Solution: Flip over the x-axis; down 3 units
Solution: 3 units right, 2 units up
Parent Graphs: Who do we know?
Who do we know?
Who do we know?
Exponential v.s. Logarithmic y 2 x y= 2 2 y y Log x x
f ( x) ( x) f ( x 2) ( x 2) 2 2 SHIFTED 2 units to the right
f ( x) x f ( x 3) x 3 SHIFTED 3 units to the left
f ( x) x f ( x) 3 x 3 SHIFTED 3 units up
f ( x) x 3 f x 3 ( ) 3 x 2 SHIFTED 2 units down
f ( x) x f (3 x) 3x Horizontally Compressed 3 times
f ( x) x 2 1 1 3 3 2 f ( x) x Horizontally EXPANDED 3 times
f ( x) x 3 3 f ( x) 3( x) 3 Vertically EXPANDED 3 times
f ( x) x 1 f ( x) 1 2 2 x Vertically Compressed 2 times
New Function Graph (or not): Rational function
Definition: Asymptote Vertical Asymptote: (imaginary) line which corresponds to the zeroes of the denominator of a rational function Horizontal Asymptote tells us where the graph will being going when x approaches infinity
(Pair Work) Sketch :
Transformation with Asymptotes VA HA
Solution for #3
Additional Example 1 Graph: f(x) = 1 x + 4 x + 4 indicates a shift 4 units left Vertical Asymptote: x = -4 No vertical shift Horizontal Asymptote: y = 0
Additional Example 2 1 Graph: f(x) = 3 x + 4 x + 4 indicates a shift 4 units left Vertical Asymptote: x = -4 3 indicates a shift 3 units down which becomes the new horizontal asymptote y = -3. Horizontal Asymptote: y = 0
How about.? 1 1 f ( x) x & f ( x) x x 1 x 1
The Vertical Asymptote is the same for both: x=1 But now you have a slant asymptote for each: y=x or y=-x Also, notice the location of the curves
What do you notice?
Happy Wed. Cell-phone in your bag Notebook Pens/Pencils Ready to learn Markers White board Agenda Rational Graphs Part 2 Long Division Quick Check this Friday
Pair Work: Copy then Sketch these four equations (10 minutes) 2 2 f ( x) x & f ( x) x x 1 x 1 2 2 f ( x) x 3 & f ( x) x 3 x 1 x 1
The Vertical Asymptote is the same for both: x=1 But now you have a slant asymptote for each: y=x and y=-x Notice the location of the curves because of the -2
What do you want to add to your notebook for these types of equations? 2 2 f ( x) x & f ( x) x x 1 x 1 2 2 f ( x) x 3 & f ( x) x 3 x 1 x 1 HINT: Tips to help you sketch these types of rational graphs
The parent function 1 x How about this one? 1 x 2
Transformations 1 2 x 2 1 x 2 1 ( x 3) 2 1 ( x 3) 2 4
Rewrite rational function standard form to transformation form
Example1) Given : x 2 8x 17 x 3 x 5 2 x 3 x 8x 17 2 ( x 3 x) x 3x Solution : 5x 17 (55 x 15 15) 2 x 5 2 x 3
Example 2) p 2 2 p p 6 20 p 6 = p 4 p 2 2 p p 4 44 p 20 6 44 p 6 ( p p p 6)( 2 2 p 4 p Check 4) 2 p ( p 6 p 20 6) 24 44 p 6 44 2 ( p 6 p ) 4 p 20 4 p 24 ( 4p 24) 44
x 1 x 3 x 2 2 x 3 2 ( x x ) 2x 2 2x 4x 4x 2x 2 2x (2x 2 x) 6 6 6x 6 6x 6 (6x 6) 1. x goes into x 3? x 2 times 2. Multiply (x-1) by x 2. 3. Subtract the binomial 4. Bring down 4x. 5. x goes into 2x 2? 2x times 6. Multiply (x-1) by 2x. 7. Subtract 8. Bring down -6. 9. x goes into 6x? 11. Subract. 6 times 10. Multiply (x-1) by 6. EXAMPLE 3 0
Example 4) Synthetic Division Divide x 4 10x 2 2x + 4 by x + 3-3 1 0-10 -2 4-3 +9 3-3 1-3 -1 1 1 x 4 10x 2 x 2x 4 x 3 3x 2 x 1 1 x 3 3
Extra: Write down your observation for the equation and its graph
Practice: sketch these functions (10 min)
9/14/17 Review Assignment Domain Range EB.. Interval of increase : decrease : f (4) f (3) f ( x) 2 x? x y intercept intercept Features of a graph (10 min.)
Domain (, 1) U [1, ) Range (, ) EB.. Interval of f f,, increase : (, 1),(1,3) decrease : (4) (3) x x 1,1 intercept y y (3, ) 2 (6 is excluded ) f( x) 2 x? 1, 3( 1 is excluded ) x y intercept 1.8,5 none
9/14/17 Review assignment
a) f ( 2) 6, g( 3) 3 false b) g( 4) not a solution c) f ( 3) 3, y int. of g(x) 2 true d ) x 2, 6
9/14/17 review assignment
Solutions for Part c) Solutions for Part d)
9/14/17 review assignment handout E) Sketch : 2 2 1) y 4 2) y 4 2 x 3 ( x 3) 2 2 3) y 4 4) y 3 x 3 x 3 1 1 5) y x 2 6) y x 1 x 3 x 3 7) y x 2 8) y 9) y x 3 x 3 x 3 2 3 x 4 3x 15x 18
Challenge Problems (DOK3?)
Find : w( 1) 4 f (2) f (1) End Behavior : Left Domain : (,8] Ra
Happy Friday 9/15 In your seats when the bell rings Cellphones in bags Pencils/Pens notebooks out Be ready Quick Check (25 minutes) Finding domains given equations (not graphs) Reflection and Practice
Domain Range EB.. Interval of increase : decrease : f ( 4) f (3) f ( x) 2 x? x intercept y intercept NOW, answer these questions with your partner(s) 7 minutes
D : (,0) U(0, ) E. B : left x, y R : (, 2] right x, y 1 Interval of increase : (, 2) U( 1,0) U(1, 2) decrease : ( 2, 1) U(0,1) U(2, 3) f( 4) 1 f(3) 0 f ( x) 2 x not visibleon the graph but est. 6(?) x intercept est.( 4.5, 0) and(0.5,0) and(3,0) y intercept (0, 2) is an open pt so none
y x Find the intervals of increase and decrease. x
y x Or where (x) does the graph go up? Where (x) does it go down? -6-5 -4-3 -2-1 1 2 3 4 5 6 x
Happy Monday In your seat Cell-phone in your bag Notebook Pencils & pens Markers White board Agenda: Warm-up Practice Domain (Algebraic) Operations of functions Self correct QuickCheck QuickCheck 2 this Thursday
Warm-up 10 minutes
Warm-up #2 Given the functions below : 4 6 5 5 h( 10) g(15) f ( x) 20 g(x) 10 2 2 f ( x) x 20 and g( x) x 20 and h( x) x 2x x intercept for g(x) y intercept for g( x)
Solution h( 10) 80 g(15) 2 f ( x) 20 x 50 g(x) 10 x x intercept for g(x) y intercept for g( x) 50 x y 20 3 25 3
Extra Practice
Extra Practice
Recall domain meant the set of values you plug in for x. For the functions we usually deal with, there are two "illegals": 1. You can't divide by zero (denominator bottom of a fraction can't be zero) 2. You can't take the square root or any even root of a negative number When you are asked to find the domain of a function, you can use any value for x as long as the value won't create an "illegal" situation.
Find the domain for the following functions: f x 2x 1 Note: There is nothing wrong with the top = 0 just means the fraction = 0 Since no matter what value you choose for x, you won't be dividing by zero or square rooting a negative number, you can use anything you want so we say the answer is: All real numbers x. g x illegal if this is zero x x 3 2 If you choose x = 2, the denominator will be 2 2 = 0 which is illegal because you can't divide by zero. The answer then is: All real numbers x but x 2. means does not equal
Let's find the domain of this one: h x x 4 Can't be negative so must be 0 x 4 0 solve this x 4 We have to be careful what x's we use so that the second "illegal" of square rooting a negative doesn't happen. This means the "stuff" under the square root must be greater than or equal to zero (maths way of saying "not negative"). So the answer is: All real numbers x such that x 4
Domain for x 2 h x 3 9 Can't be negative so must be 0 x x 2 2 9 0 9 x x x 2 9 3 3and x 3
9/14/17 review assignment
Solutions for Part d)
Quick Check 1 Self-Evaluation Make Correction of your answers What types of mistakes did you make? CO conceptual errors, i.e. you did not remember the definition for domain CA calculation error, careless mistakes Assign a grade as student (yourself): 1 only some parts of #1, 2, and 3 correct 2 - #1, 2 correct with 1 mistake; #3 correct with no mistakes 3 - #1, 2 correct and a part of #3 correct 4 - #1, 2, and 3 correct with a CA mistake Turn the quick check in
Happy Tuesday In your seat Cell-phone in your bag Notebook out Pencils & pens Markers White board Agenda: Operations of Functions: substitution & composition Practice QuickCheck 2 this Thursday
Concept Category 1 Part 3 Function Operations Graphically Algebraically
We will be going over Substitution (a type of composition) Composition of functions (still substitution really) Adding/Subtracting functions Multiplying/Divide functions Inverse function
Find f(2k-1) if f x 2x 2 3x 6 Substitution 2k 1 2k 1 2k 1 f This means to find the function f and instead of having an x in it, put a 2k-1 in it. So let s take the function above and make brackets everywhere the x was and in its place, put in a 2k. 2 2 3 6 2 k k k 2 4 4 1 3 2 1 6 2 8k 14k 11 Don t forget order of operations---powers, then multiplication, finally addition & subtraction
* Substitution Practice
Composition of Functions Ex 1) The Alternative Symbol for Composition : f ( g(x) ) f g(x)
Practice Now
Happy Wednesday In your seat Cellphone in your bag Notebook, Pens/Pencils out Agenda for today Active Practice for CC1 Part 1, 2, and 3 (substitution & composition)
Always have your notes with you Anything missing? Ask a peer first what did you miss? Do you all have the same question? Still stuck? Then ask the teacher * Might need to update notes*
9/19 th Active Practice For the graph ( it ' s not a function) : a) Domain? b) Range? c) End Behavior? d ) Intervals of increase? Decrease? e) x intercept(s)? f ) y intercept(s)? 1 g) f (3) 3 f (0) 3 Sketch a graph with these features:
Sketch : a f x x g x f x h x x b x h x 2 ) ( ) ( ) 2 ( 3) 6 b) ( ) ( ) 3 ( 4) 1 1 1 c) t( x) u( x) 2 t( x 3) 4 d) v(x) c( x) 3 v( x 2) 2 2 x x x 1 e) z(x) 2 n( x) z( x 3) 1 2 a) f g( x) b) g f ( x) c) g g(x) d) f g f (1)
Given the functions below: a) 3 g(2) 3 h(4) f( 5) b) g f h( 1) c) find the x intercept for h(x) Perform the operations:
Solution: D (, 4]U ( 2, ) R (, ) EB Left : x, y Right : x, y IInc (, 4) U ( 2,1) U (6, ) IDec (1, 2) U (2,6) x int : ( 0.5,0),(5,0), and (7,0) y int :(0,1) 13 3
2 4 a) D (, ) b) 6 D (,3) (3, ) x 2 2 3 ( x 3) 4 2 1 c) x 12x 42 d (, ) d) 2
a) 16 b)6 c) (2,0)
a) g(6) b) f(3) c) x 2, 5, 7 d)[1, 2] U[5,7] e) (2,5) U(7,8)
Extra Practice textbook pg156 On substitution (composition):
szxa On features of a graph (pg 182)