Mr. O'Brian,
Overall the class believed that the homework went well and was a fair amount of problems. Some sections that the class had trouble with was pg. 49 (65, 75-78) The class seemed to not clearly understand domain. More importantly the class did not know how to find the domain easily, we were wondering if there was a equation to help solve the problem. On problems, 75-78, the class also wondered if there was a equation to find the solution, instead of simply guess and check. One of our class questions for the quiz next class is if we are able to have a note card to write the formulas on?
Hope your feeling better,
Katharine
Note: Tyler will be the next class scribe
Subscribe to:
Post Comments (Atom)
Thanks for taking the lead, Katharine! Who is the scribe for next class?
ReplyDeleteLet's see...
Domain will be on the quiz tomorrow, so let's make sure everyone feels comfortable with it. Very simply, the domain of a function is the set of input values of the function. Here is a more detailed explanation:
http://www.purplemath.com/modules/fcns2.htm
When given a function like f(x) = 1/x - 3/(x+2), almost any input works in the function- except 0 and -2. The reason for this is that these two values "break the function" by leading to division by zero.
In the case of domain, we can generalize from our understanding to form two "rules". The first is that any values of x which lead to division by zero must be eliminated, so take the denominators of any fractions in the function and set them to zero to find the values that don't work- the remaining real numbers form the domain of the function. The second is that any values of x which lead to a square root of a negative number must be eliminated, so take the radicand (thing under the radical) and make an inequality to find the numbers that work (or do not work). So, finding the domain of a function like g(x) = sqrt(10 – x) consists of solving the inequality 10 – x ≥ 0.
We will discuss 75-78 further in class, so they will not be on the quiz. While we will form some generalizations of how to handle a problem like this (strategies like looking at the problem numerically and graphically as well as algebraically), we will not come up with "rules". This is one of the key differences between this course and Algebra 2. In Algebra 2, much of the work consists of learning algorithms and applying them. While we use those algorithms this year, there is much more thinking required in their application.
Finally, I can't think of a formula that you would need on a notecard, but if you can give me a specific case, I can give you a way to understand the formula and not memorize it. If I can't do so, I will let you bring it in on a notecard. So, for now, no notecard and no memorization required!
Oh yeah, one other thing. Reading an explanation can help, but there's no replacing a good face to face chat. So, if you'd like to discuss any of the above in person, stop by during study hall, or visit another math teacher during Math Directed Study:
ReplyDeletehttp://math-ob.wikispaces.com/Mr.+O%27Brien%27s+Schedule#toc3
Hi Mr. O'Brian,
ReplyDeleteA specific case of a formula that I was considering was something along the lines of the distance formula: sqrt (x2-x1)^2+(y2-y1)^2. Others were the mid point, parallel line, and the perpendicular line formula. However, like you said if we spend some time to understand the formula next class before the quiz, there would not be a note card needed.
Hi Katharine-
ReplyDeleteThanks- who's the scribe for today?
The distance formula is just Pythagoras' Theorem- a^2 + b^2 = c^2 where a and b are legs of a right triangle, and c is the hypotenuse with the hypotenuse solved for. The x2 - x1 is the length of one leg and the y2 - y1 is the length of the other. When you plot the points, you see it nicely.
As for midpoint, if you want to find a number halfway between two others, you average them- e.g. the number halfway between -6 and 10 is (-6 + 10)/2 = 2. So, to find the midpoint between two coordinates, you average the x's and average the y's.
For parallel lines, since they have the same direction, they have the same slope- i.e. slopes are equal.
For perpendicular lines, a sketch shows that if one slope is positive, the other must be negative. As for the reciprocal part, the easiest way to remember this is again with a sketch- draw a slope triangle on each line, and you'll see quite clearly that the change in x and change in y are reversed.
Let me know if this helps.
Good questions!
Sorry- I see that Tyler is the scribe!
ReplyDelete