Official December SAT II Math 2 Thread

<p>It was I only. III's graphs of its inverses are not the same; therefore, they are not equal.</p>

<p>III's graphs are definetely equal. You can graph it yourself using parametrics</p>

<p>the graph y=1/(x-1)+1 has vertical asymptote at x=1 and horizontal asymptote at y=1; when you switch the x's and y's and solve for y, you will get the same equation y=1/(x-1)+1 and the same asymptotes (since both are at 1, switching x and y does not matter).</p>

<p>Do you guys think the curve this year is going to be more generous, say a 42 instead of a 43 for 800?</p>

<p>I just remembered one question I wasn't sure of...I think it was like:
If a, b, and c are real numbers and x=ax^(2)+bx+c, then what would be the maximum number of zeros? I put 2. Anyone else?</p>

<p>i vaguely remember that question..i don't really remember if it asked for the maximum number of zeros or not, but i think i put 2 as well</p>

<p>yes x², 2 zeros.</p>

<p>i thought 2 was minimum and there was no max.........</p>

<p>i thought thats what sparknotes said.. hm.</p>

<p>ah crapoo maybe im wrong.</p>

<p>But if you think about the equation x=ax^(2)+bx+c, x (or y=x) is a straight line that intersects the origin and a^(2)+bx+c is some sort of parabola. Now the maximum number of intersections between these two graphs is two (this is how I came up with the answer).</p>

<p>dinosaurrawr, 2 can't be the minimum...there could be one zero, if the polynomial is a perfect square, etc.</p>

<p>yeah i guess, i guess i was thinking the opposite.
oh stab me.</p>

<p>i put "can't be determined"
eh crap.</p>

<p>A quadratic can intersect the x-axis zero (y=x2+1), one (y=x2), or two (y=x2-1) times.</p>

<p>so maybe.. "can't be determined"</p>

<p>There was a question like "which of the following has exactly two roots?"</p>

<p>I put y = (x2+1)(x2-1) <-- pretend those 2's are exponents</p>

<p>yeah, thatz what I put, because x=1, -1 are the two roots. The other choices had no, one, or more than two roots</p>

<p>anyone else concur?</p>

<p>just graph it on your calculator jeez and yes its right</p>