TO ALL LOVERS OF ALGEBRA (yes, that's you guys!)
I cannot over-emphasize the importance of ALL of the homework problems I assign in this unit. If you are having difficulty, please ask a question on the blog... DO NOT WAIT FOR CLASS TO ASK QUESTIONS!!!!
YIPPEE!! Unit 9 is up and running on MathChamber for your viewing pleasure.
Please take advantage of the VIDEO TUTORS in this unit
You can use a graphing calculator to support your work
You should be able to answer many of these questions without generating a specific graph...
... for example #34-39 & #44-47 should be "at-a-glance" answers
BOTH ASSIGNMENTS ARE DUE ON THURSDAY!!!!
hw #9-1
pg 538-540,#5-10,15-20, 34-39,41-47,52-57
hw #9-2
pg 544-546 #11,15,16-19,36-40,46-47
Are we supposed to do all of this by Wednesday???
ReplyDeleteI was confused on how to find the domain of a function/graph
ReplyDeleteYUP!! It's really not that much... many of the answers are true "quickies" and you can use the graphing calculator.
ReplyDeleteGentle reminder: I tried to talk YOU out of this in September, but did YOU listen?? NOOOOOO!!!! It's time to "pay the piper" my friend... get to work!
i am confused on number 40 and dont know where to start
ReplyDeleteDomain is the possible 'x' values and range is the possible 'y' values for a given function. We discussed domain and range several units ago... nothing has changed.
ReplyDeleteTake a look at problem 2 on page 535... that should answer your questions.
Parabolas "stretch" across the entire coordinate plane just as linear functions do... from a vertical perspective, every quadratic function (parabola) has a vertex, which is either a minimum point ('a' is positive) or maximum point ('a' is negative).
This means that the domain for a quadratic function will ALWAYS BE 'all real numbers' and the range will be restricted to y>=some# or y<=some#, depending on the nature of the quadratic ('open up' or 'open down').
Did this help?
To RE-answer your first question, you should have this assignment done by tonight so that you can ask questions and get them answered by Wednesday night... so get to work, you're behind schedule!!
ReplyDeleteyou didnt answer my question on number 40 yet!!!!!!!!!!!!!!
ReplyDeleteIf you are talking about problem #40 on page 539, I SPECIFICALLY DID NOT ASSIGN THAT PROBLEM. Fear not, we will cover that type of problem in our next class.
ReplyDeleteIf you are talking about problem #40 on page 545, you should be able to figure that one out, but don't spend too much time on it.
Wear comfortable clothes on Thursday... we're going outside if weather permits.
Dear Impatient... and just how much are you paying for this 24x7 on-line tutoring resource?????????????????????
ReplyDeleteyay...outside sounds fun
ReplyDeleteWhat? "yay...outside sounds fun"?
ReplyDeleteOh well =)
Wait WHAT?!? WHY ARE WE GOING OUTSIDE?!?
ReplyDeleteI just got it =D
Would the answer to #5 be "the vertex of the parabola is the minimum height when a is a positive number, and the vertex of a parabola is the maximum height when a is a negative number"?
ReplyDeleteFor #10, is there another way to solve -4x^2 other than making a table of values? If there is, sorry I'm forgetful from class!
ReplyDeleteFor #41-43 do we need to copy the graph onto our sheet of paper from the graphing calculator? Or do we just need to identify the vertex and axis of symmetry?
ReplyDeleteIf something asks for the axis of symmetry, and the line of symmetry is on the y axis, do we just put "y-axis"? And if it is not, do we need to come up with an equation for the line?
ReplyDeleteI don't understand what they mean by "|a| has the greatest value" or "|a| has the least value" in #46 & 47 on pg 539.
ReplyDelete*Gasp*
ReplyDeleteMULTIPLE CHOICE QUESTIONS!
SO GENEROUS!
How do you use factoring to solve #56, if it can't further be factored?
ReplyDeleteDear MinMax#5... Yup!
ReplyDeleteFor 538#10 & #15, I am expecting you to use a table of values (5 'x' values is enough). That said, I would also expect that you would have a general idea of skinny vs. wide, open-up vs. open-down and (especially) in the case of a y=ax^2 (i.e. 'b' and 'c' both equal zero), you should know EXACTLY where the vertex is located.
For #41-43, at this point in your mathematical life, a little "napkin sketch" would be nice... but yes, it is just asking for the vertex and axis of symmetry.
ReplyDelete|a| means the absolute value of a, right??
ReplyDeleteQ. Which parabola is wider, y=2x^2 or y=-2x^2??
A. Neither.
Remember, the 'a' value controls "skinny/wide"... in effect, this means that 'a' values that are opposites "clue" you in to the width of the parabola. Clearly, line M is the widest parabola, and just as clearly, the 'a' value must be negative (i.e. open-down). So the absolute value of the a-value of line M must have the greatest value... ca-peesh??
Dear Short of Breath (GASP)!
ReplyDeleteFriendly reminder: YOU WILL get multiple choice questions wrong on the ASK test and the Algebra EOC test... MCQ's are notorius for "teaser" answer choices that make you "leap before you look"... so be careful, cowboy... or cowgirl.
Dear #56,
ReplyDeleteFactoring is FUN & EASY... you are looking for two numbers with a PRODUCT of -48 and a SUM of -22.
You simply list the possible dimensions as the BINOMIAL FACTORS of the given trinomial.
Surely, you had FUN FACTORING, yes?
Oh, alright! (For all the above)
ReplyDeleteand YEE-HAW!!
Wait, -b/2a solves the x-coordinate of the line of symmetry?
ReplyDeleteAm I doing something wrong? On #11 on pg 544 I got x= 1 1/3.. And then I have to square it? And I get a really long decimal?
ReplyDeleteAnd then when you solve the function with the solution of -b/2a do you get the y coordinate?
ReplyDeleteI've got the vertex for #15, how do I figure out the parabola? (VOCABULARY!!)
ReplyDeleteFor #s like 16 and 19 I don't understand what the 'x' after b does to the parabola!
ReplyDeleteDear Waiter,
ReplyDeleteYup... -b/2a serves a dual purpose.
x=-b/2a is the equation for the (always vertical) axis of symmetry (sorry, I called it a "line of symmetry" in class today... "axis of symmetry" is the correct terminology).
Since the vertex ALWAYS lies on the axis of symmetry, -b/2a clearly delivers the x-coordinate value as well... DOUBLE YOUR PLEASURE with -b/2a!! YEE-HAW!!
What changes the parabola's axis of symmetry's location, on the left or right of the y-axis?
ReplyDeleteDear #11, You are right! You ARE doing something WRONG! What was your a? b? Show me your work... 11/3 is no good.
ReplyDeleteOhhhh... Wait.. Is it 1 1/2? Not 1 1/3? I made a mistake in simplifying -(-9/6).. If that is right.
ReplyDeleteFor #40, I-- Don't know!
ReplyDeleteFor #47, can I just make an area model? Or do I have to list my steps?
ReplyDeleteFor 544#15, you are ONLY being asked for the vertex and the axis symmetry.
ReplyDeleteFriendly reminder: When you finish answering a question, go on to the next one.
You are NOT EXPECTED to be able to fully graph quadratic functions like #15 without a table of values OR a graphing calculator. That said, YOU SHOULD be able to locate the vertex (-b/2a is not rocket-science), recognize open-up vs open-down, and have a GENERAL IDEA of skinny vs. wide... ca-peesh??
Oh okay I got it! Sorry, I misread the problem.
ReplyDeleteWARNING!!! We focused on the fact that the b value impacts the left/right movement of the parabola... BUT (and that's 'BUT' with one 't'), notice what happens to -b/2a when BOTH 'a' and 'b' are negative (as in graph C in #16-19). Funky stuff, eh?
ReplyDeleteThis is why I showed you the FUNCTION FLYER graphing calculator. Play with various 'b' values and see for yourself. Use both +1 and -1 for an 'a' value.
Isn't algebra simply fascinating??!!
Fascinating but confusing!!
ReplyDeleteFor 544#11, -b/2a=-(-9/6)=3/2=1.5
ReplyDeleteI don't know where 11/2 or 11/3 enters into the picture?
For #47, I would want you to be able to ARTICULATE that by using the area model, you account for each of the six multiplications that must take place.
ReplyDeleteI don't like the "justify each step" directive... they could have simply asked you to explain why your method works.
Carry on.
545#40, I'll save for class discussion. It's easier than it might at first appear.
ReplyDeleteOH sorry I don't mean 11/3 I mean 1 & 1/3 and 1 & 1/2!! I just spaced it out like this: 1 1/3
ReplyDeleteOkay..
No pain, no gain. Confusion is pain for the brain. Confusion is good.
ReplyDeleteOkey dokey!
ReplyDeleteI will try to solve the problems you answered tomorrow.
Good Night =)
Yup... 1 1/2 = 1.5... we're good!
ReplyDeleteOkay! =)
ReplyDeleteI don't know how to justify the steps for #47.. I don't know why the area model works!
ReplyDeleteIt's as simple as I stated above... it ACCOUNTS for all six of the multiplication problems that must be then summed. The area model is simply a structured "graphic organizer" to layout the individual products (resulting terms) that then need to be checked for like terms.
ReplyDeleteIt can also be thought of as an alternate way to view the distribution property.
That;s all, don't sweat the explanation, as long as you can do this problem below:
(3x^3-4x^2+x-7)(4x^4-5x^3+7x^2-4x+3)
Distribute or AREA MODEL... you tell me!
(P.S. You don't have to do it)
I don't understand 44, 45, 46, 47! Well, I understand it but I'm not sure how to do it?
ReplyDeletefor 44-47 a is the variable a in each graph to the right. Variable a is the one that goes next to x^2 in the equations used to represent the graphs to the right
ReplyDeleteYes, 'a' is the coefficient of a quadratic function in standard form... y=ax^2+bx+c, right?
ReplyDeleteI also answered a question like this up above last night (see 7:48pm)
Okey dokey! About the area model!
ReplyDelete