<p>I'm fine on the rest of the SAT math section, but when I get to functions it's like I'm looking at another language.. I have NO idea what they mean or are trying to get me to do. Are there any sites out there than can explain them to me? It's the only thing keeping me from 800 in math (I would think)</p>
<p>+1 here man. functions own 9 out of 10 times. if you find any good sources for help, don't be a stranger :)</p>
<p>What kind of functions are giving you the problem?
Put an ex. on here and how you tried to solve it.</p>
<p>functions are the easiest for me there just not taught much.</p>
<p>well I was never taught them...so I have no idea what's going on with them..like Question 18 on pg 400 in Blue Book...no idea what to do</p>
<p>that question isn't even asking that much about the function. </p>
<p>Think about it. If (a,b) are part of the shaded regions, then:</p>
<p>I is right because a has to be less or equal to 4 or it would be out side the 4.</p>
<p>II would be wrong since b is greater than a in the point (2.5, 3)</p>
<p>and III is true because if a=4 then the y would equal the highest point that b could be. what ever y is, that is the highest point b could be (and less).</p>
<p>The function in this problem just set one of the limits for the shaded region.</p>
<p>If you don't know functions, there is no way your close to a 800. What math classes have you taken?</p>
<p>^^ no, I'm not close to 800 yet (first practise test i ever took, cold turkey was 580-660), but all the questions that I missed I just missed because of a lack of practise in math over the summer... I could do all of them them after thinking about them for a while after looking over the test EXCEPT the ones involving functions </p>
<p>My math classes have been pretty bad. Grade 8-9 was awful because I was going to a crappy school, they were both pretty much a vertbatim repeat of grade 7. Then last year I went to a pre-IB program and took Math 10 Enriched (so the province's math 10 curriculum in 2 terms rather than 3 and the first term of Math 11 IB in the third term). We just quickly went over functions, maybe for 2 classes, and didn't teach them as much as the SAT expects you to know</p>
<p>Could you post some more examples on where you are having trouble? It is difficult to explain what a function is without getting technical when there are no examples to illustrate.</p>
<p>well in the blue book from page 256 to 262 are all about functions, I don't understand any of it except the one with the symbol..more precise? Uhh, well on 259 at the bottom it says "Which is a graph of y= -x^2-1?" and then shows a bunch of different graphs. The only way I would be able to do it was to punch the function into my calculator and make it graph it out..</p>
<p>I don't have the book, but with reference to your question, you just have to know basic scaling and shifting. Let me list this.</p>
<p>f(x+a), shifts the graph to the left 'a' units.
f(x-a), shifts the graph to the right 'a' units.
f(x) + a, shifts the graph up 'a' units.
f(x) - a, shifts the graphn down 'a' units.
-f(x), flips the graph over the x-axis. </p>
<p>By combining these rules, you can spot these questions in a few seconds and will save you a lot of time. </p>
<p>For f(x) = -x^2-1, think of this function as an extension of f(x) = x^2. This function is at a parabola at the origin. Have this picture in your head and then just apply the above rules, as necessary, to the function step by step. </p>
<p>So, first they projected x^2 across the x-axis. (this is -x^2) So now it is a parabola, still at the origin, but upside down. Then this graph is shifted down 1 unit. (since it is now -x^2-1) This just means that every 'y point' of the graph is lowered by 1. So instead of a graph with maximum at origin, it is now at (0,-1) with the remaining graph drawn accordingly. </p>
<p>Just take it in steps, these type of problems are really freebies on the SAT.</p>
<p>f(x+a), shifts the graph to the left 'a' units.
f(x-a), shifts the graph to the right 'a' units.
f(x) + a, shifts the graph up 'a' units.
f(x) - a, shifts the graphn down 'a' units.
-f(x), flips the graph over the x-axis. </p>
<p>By combining these rules, you can spot these questions in a few seconds and will save you a lot of time.</p>
<p>For f(x) = -x^2-1, think of this function as an extension of f(x) = x^2. This function is at a parabola at the origin. Have this picture in your head and then just apply the above rules, as necessary, to the function step by step.</p>
<p>So, first they projected x^2 across the x-axis. (this is -x^2) So now it is a parabola, still at the origin, but upside down. Then this graph is shifted down 1 unit. (since it is now -x^2-1) This just means that every 'y point' of the graph is lowered by 1. So instead of a graph with maximum at origin, it is now at (0,-1) with the remaining graph drawn accordingly.</p>
<hr>
<p>thanks a lot, those rules are VERY helpful now that I better understand what's going on.. Just to clear something up, to see if I understand, if it says "f(x) = (x+1)^2" it would be a parabola that's been shifted to the left 1 unit?</p>
<p>"if it says "f(x) = (x+1)^2" it would be a parabola that's been shifted to the left 1 unit?"</p>
<p>Exactly.</p>
<p>ok, one more thing..</p>
<p>is there a difference between f(x)=x^2+1 and f(x)=(x^2+1)</p>
<p>"is there a difference between f(x)=x^2+1 and f(x)=(x^2+1)"</p>
<p>No.</p>
<p>so that one would shift the graph up 1 unit?</p>
<p>Right. Every 'y value ' of the function will be increased by 1.</p>
<p>Don't try to memorize things like this, they are very easy to actually understand. Take a look at graphs and try to think about what happens if everything had +1 to it, or whatever. Try to think of real examples, etc.</p>