They do what they do.
Thinking about schools and peers and parent-child attachments....I came across one of my favorite posts .
I don't understand the explanation of this problem.The answer is D, and the first line of the explanation says:"The function with equationy=(-x)^2+1￼ and the function with equationy=|x^2+1| each have a minimum value of 1 when x=0, but the function graphed does not have a minimum value of 1, so these options cannot be correct.￼"I don't see what x=0 has to do with the minimum. Do I not understand what the minimum is?
Both of those functions have a minimum value of 1, and that minimum occurs at x=0. The graphed function has minima of 0, at some positive x and some negative x.
Both of those functions have a minimum value of 1, and that minimum occurs at x=0.oh!Thank you so much.
I would spend time thinking about various graphs. What does the graphy = (ax^2 + b) + clook like for various values of a,b,c?When does it have a y intercept? An x intercept? What are the extrema? What does an absolute value sign change, if you had to graph it?
I would spend time thinking about various graphs.right - the issue here is that I have never taken algebra 2. Or, rather, I seem to have taken parts of it, as I've worked my way through all of Saxon's Algebra 2, which is the 2nd book in a 3-book cycle.I'm a complete novice on graphing functions. I've spent lots of time factoring quadratic equations, completing the square, and deriving the quadratic formula. But none of that knowledge is particularly well connected to graphing.That's why Dolciani's chapter on quadratic functions is up next.
Same issue with probability. I've worked my way through Dolciani's chapter on it & have not had a chance to do it again -- or to work my way through other textbook presentations. I've gone through several SAT review lessons on probability, which taught me that you cannot learn a subject by reviewing it for a test.
This is an Alg I problem...or at least the way my district ran the Integrated Alg.I course last year.Should you not know a lot about parabolas, it's easy to figure out the right answer by going through the answers and plugging in a few different values of x to see if they lie on the graph.
Interesting.I'm pretty sure SAT considers this algebra 2 -- in any event, this material is in Dolciani's Algebra 2 book.
I used the Arlington Algebra Project Material Unit 5 to get the parabola definitions and properties quickly into my son. If you're pressed for time, it's a good condensation. If you're not, working the Dolciani exercises is good.
I'm sure these problems are in algebra 2 books, as well, but in terms of *content*, and at the risk of offending people, there is nothing in any of the problems you've presented that required knowledge past pre-algebra. Simple inspection would give you the right answer here--insert value for x, read off value for y.None of these problems required any formula past the pythagorean theorem. None required any algebra 2 whatsoever. None required knowing any relationships about ratios of radii, or bizarre slanted triangle heights. It didn't require knowing the hypergeometric distribution, or what a bernoulli trial is.What they required past pre-algebra was more mathematical maturity than most students have in pre algebra, and in fact, more maturity than most students have even in algebra 2.The difference is that all you needed to do was think in a step by step fashion about what you were given. This is exceedingly difficult! It isn't taught in a few months or even a couple of years. That's why cramming the sections of Dolciani can only do so much.What you are trying to figure out in record time is how to approach thinking about problems symbolically. But that's really where you need to start, and all of the rest of the studying is just other ways to try and connect the dots so you can start thinking symbolically.
I look at that graph and I think, "Huh. If that middle hump were flipped upside down, this would look like a parabola." And then I think, "In fact, if you took a parabola that dipped below the x-axis and then wrapped it inside an absolute value function, all of the negative stuff would flip upside down and you'd end up with a graph like this."So that tells me I am looking for a quadratic function inside absolute value bars. Then I (mentally) sketch the parabola by flipping the hump back over and I realize I'm looking at x^2 - 1.
Post a Comment