I have reviewed EACH AND EVERY question in the Unit Review (excluding 9-7 thru 9-8) and the Chapter Test (excluding #20-22). I feel every question is fair game for the test, including the word problems and vertical motion problems.
P.S. I'll be home late-ish on Thursday (more math debates), so you'll need to help each other with questions and answers... please do.
how are you trying to get started with Review #36?
ReplyDeleteI don't know the answer to #3 on pg 589.
ReplyDeleteFor #11 on pg 590, I tried to graph it and made a table of values. I substituted in 0 for x and got the y coordinate to be -3.. Is that even possible? My vertex is at (2,3)..
ReplyDeleteAnd all of my 'y' coordinates were under the vertex, even though the parabola is supposed to open up because 'a' is positive..
ReplyDeleteis at weird that I find it easiest to make a table of values...
ReplyDeleteFor #11, that's a good y-intercept, but your vertex is off, are you substituting your negatives using parentheses?????
ReplyDeleteIt's good that you can make a TOV, but an algebra-tician needs to factor, complete-the-square and use the QF.
ReplyDeleteI'll be leaving in a few, be back ??? 8-10 ???
ReplyDeleteI'll be in school 7am tomorrow...
For #11, I'll see..
ReplyDeleteand could I come in at 7?
for 589#3, the DISCRIMINANT is the name we give to that magical value that lives deep within the QF, hidden within the square root symbol... b^2-4ac.
ReplyDeleteIf the discriminant is positive, the QF will generate two numbers (because of the +/- effect)
If the discriminant is zero, the QF will only generate one number, since the +/- zero is not going to change '-b', right?
If the discriminant is negative, well STOP right there, since we can't square root a negative number, the QF won't generate ANY numbers at all.
... and that's what it's all about!!
I'll be there at 7am, visitors welcome.
ReplyDeleteDo we only do the review sections up to 9-6?
ReplyDeleteI will come in at 7 also
ReplyDeleteOkey dokey! And got it!
ReplyDeleteDid I do something wrong on #1 on the Chapter Test? I got the axis of symmetry as 7/6, and my vertex as -2.92.. But when I plug in 1 into the table of values, it came out as -4! And the parabola is supposed to open up..!
ReplyDeleteI'm confused on all of the graphing problems.. Why is it so hard? >:(
ReplyDeleteWait, if you get 2 answers for a vertical motion problem, which one is correct? The one that is not negative?
ReplyDeleteAnd I am coming tomorrow @ 7!
On the test, will we have to create an equation from a vertical motion word problem including velocity? Do we have to figure out the velocity or will it be given to us?
ReplyDeleteAlso, do we have to map out the "speed of gravity" and how much it decreases per second? Like Karel's rock problem? Because I do not understand those..
I will try to check the blog in the morning.. I have to go to bed soon..
ReplyDeleteOh & please answer the questions!
ReplyDelete& for the 3rd time, I am coming at 7 =)
In ChapTest#1, b=0.
ReplyDeletey=3x^2-7 can be re-written as: y=3x^2+0x-7
so:
a=3
b=0
c=-7
If someone drops a rock, and you want to know how long it takes to fall to ground level, and the resulting quadratic gives you two solutions (one positive, one negative) the positive solution "counts" and the negative solution is "extraneous."
ReplyDeleteIf the negative solution "counted" time would have to start moving backward when the rock was dropped. Karel is a special guy, but I don't think he has those kinds of magical powers!
The test is on 9-1 thru 9-6.
ReplyDeleteOhh.. For #1, right. I used c as b by accident..
ReplyDeleteAnd okay!