Midterm Exam Review

Jan 15th, 2011 by corricelli

Please use this to post questions/concerns about our upcoming Midterm Exam!

Happy Blogging,
Mrs. Corricelli

20 Responses to “Midterm Exam Review”

CLVelez, math Tutor says:

January 18, 2011 at 12:38 pm

Mrs. Corricelli Hello!
Due to the last 2 snow days and Martin Luther King Junior’s B-day. Can you please let me know, what is the new day that the Algebra II midterm will take place? Thanks!
exam will

Reply
- corricelli says:
  
  January 18, 2011 at 2:23 pm
  
  This Friday!
  
  Thanks,
  Mrs. Corricelli
  
  Reply
Kate says:

January 18, 2011 at 8:43 pm

hi mrs. corricelli
on the midterm packet i keep getting the questions for #56 wrong
how exactly should i start those problems?

Reply
Lauren says:

January 19, 2011 at 4:39 pm

Hi Mrs. Correcelli,
I’m having trouble with #48 in the midterm review Packet and was wondering how should I start solving the the problem, I looked through my notes but am still getting a bit confused
Thanks!

Reply
- corricelli says:
  
  January 19, 2011 at 9:30 pm
  
  Hello Lauren,
  
  Could you provide the problem here? Please?
  
  Thanks,
  Mrs. Corricelli
  
  Reply
Lilah Tracy says:

January 19, 2011 at 8:21 pm

Hi Mrs Corricelli!!
I have look in my book and my notes and i still can’t seem to get this question right! its # 20 in the packet. I think I’m getting confused by the two different f of x equations! i could really use you help! Thanks!

f(-3). f(x)= { -x2+2x
-2x+3
x1

Reply
- corricelli says:
  
  January 19, 2011 at 9:33 pm
  
  f(-3) means that, in f(x), wherever you see x, put in -3.
  
  I had trouble reading your post, so I’ll make up my own f(x) below.
  
  So, if f(x) = -x^2 + 2x, then f(-3) = -(-3)^2 + 2(-3) = -9 + -6 = -15.
  
  Note that the x is squared, and the negative is outside.
  
  If instead, f(x) = (-x)^2 + 2x, then f(-3) = (–3)^2 + 2(-3) = 9 + -6 = 3.
  
  I like to use parentheses around the x value so I do not make mistakes with order of operations, signs, etc.
  
  Let me know – hope this helped.
  Mrs. Corricelli
  
  Reply
Kate says:

January 20, 2011 at 9:01 pm

does anyone know if Direct Variation problems are on the midterm?? i didn’t write anything down from the review today but did i miss it?

Reply
- corricelli says:
  
  January 20, 2011 at 10:10 pm
  
  Kate,
  
  Indirectly – yes… (Sorry for the play on words – couldn’t resist)
  
  Anyhow, that problem you asked in class when you needed to get a for the parabola and you were given the vertex and another point on the parabola is a very similar question…
  
  Also, the whole: find an equation for a line parallel to y=5x + 2 that goes through (4, 2) is another version of “direct variation”… just with a non-zero y-intercept. In other words, you also need to put in x and y to get b…
  
  So, not
  directly
  
  … But still the same concepts will be assessed.
  
  Thanks,
  Mrs. Corricelli
  
  Reply
Jordan HUhtanen says:

January 20, 2011 at 9:36 pm

Hi Mrs. Corricelli
I was wondering if on the midterm for the quadratic formula, if we use Owen’s Program, will we need to show our work still on the test?

Reply
- corricelli says:
  
  January 20, 2011 at 10:06 pm
  
  Hello Jordan,
  
  You need to at least write the formula. (Which is really a non-issue once it is coded into the calculator!) One word of caution, though – if you over-rely on the quadratic formula you will miss details…
  
  For example, your solution will need to be given as an exact value (reduced square roots, and the like…) However, it will shorten your work on integer solution problems AND it is a powerful way to check your work… Use it NOW – while you are studying – so you are not surprised by anything on the test.
  
  Take care,
  Mrs. Corricelli
  
  Reply
Aubrey Wood says:

January 20, 2011 at 9:53 pm

Kate;
I don’t know if I wrote it down correctly or not but according to what I took down today, there isn’t. I also missed the first 9 so… I don’t know if I can answer accurately and that you should listen to me! Hopefully someone else can answer it…

Reply
Aubrey Wood says:

January 20, 2011 at 9:55 pm

Anyone! I’m having trouble starting piece-wise function graphs.
What’s the first step?

Reply
- corricelli says:
  
  January 20, 2011 at 10:19 pm
  
  Aubrey -
  
  You need to divide the plane into smaller sections and graph each curve/line in each section…
  
  See the following links:
  LINK 1 – good overview, a little boring, has graph paper for practice that you can print
  LINK 2 - textbook response
  LINK 3 – how to graph piecewise functions in your graphing calculator
  
  Hope this helps,
  Mrs. Corricelli
  
  Reply
Aubrey Wood says:

January 20, 2011 at 10:05 pm

Oh wait, never mind I get it! Sorry!

Reply
sam turco says:

January 20, 2011 at 10:27 pm

For the relation to be a function its just that none of the X’s can be the same on any of the points. In other words it has to pass the verticle line test, right?

Reply
- corricelli says:
  
  January 20, 2011 at 10:36 pm
  
  Hey Sam,
  
  Yes and yes! (Because, think about what it would mean if the x’s did repeat – then given an x, you would not know y – or given an input, you would not know what the output would be…)
  
  Thanks,
  Mrs. Corricelli
  
  Reply
sam turco says:

January 20, 2011 at 10:46 pm

ok thanks…. i have another question. you said question 18 would be where you give us slope+y-intercept to find y=mx+b but i don’t understand how you put that into standard form with that little information.

Reply
sam turco says:

January 20, 2011 at 10:49 pm

wait, is the standard form y=mx+b or is standard form axsquared+bx+c?

Reply
corricelli says:

January 20, 2011 at 11:04 pm

Sam,

Your question exposed some scary confusion…

Std form for a line: Ax + By = C
Std form for a quadratic: y = ax^2 + bx + c

Say slope is 4 and y-int is 3. Then, y=4x + 3
Then, -4x + y = 3, or equivalently, 4x – y = -3; both are in std form.

Signing off,
Mrs. Corricelli

Reply