<p>What about the question with the graph of a function that turned twice? It asked about whether a>0 or a<0. I put a>0. Is it correct?</p>
<p>Michael- 0.06 is right.</p>
<p>yep, a>0 on that one</p>
<p>Haha, I guessed on that! Yay! I narrowed it down to two choices but I couldn’t remember how to choose whether a is positive or negative. Any idea why it was so?</p>
<p>What about the question in which an unknown function was given and points (-2,2), (1,4) and another one satisfied it. We were supposed to comment on the interval of zeroes or something? What answer did that have?</p>
<p>I don’t think it was a>0.
There’s no way. The graph they give you is a -x^3.
If a>0 the curve that they give you would not match the curve you get from a>0.
If a<0, then you would be the -x^3 which was shown in the graph.</p>
<p>What was the answer to he first problem:)</p>
<p>haha because on the right side it was positive, and the left was negative.
so for example if 5 is cubed, it’s positive. but when -5 is cubed, it’s negative. the 3rd degree polynomial should be positive to get that choice. “a” decides whether the entire polynomial is pos. or neg. so a > 0 is the right choice</p>
<p>TheMathNerd, the graph was one of the positive function f(x) = x^3</p>
<p>The left was positive and the right was negative…</p>
<p>how big do you guys think the curve will be? if i got say 5-7 wrong and omitted 5, what would that probably come out to? i know on the practice test i took from the college board prep book, you could get literally 7 wrong and still have an 800…</p>
<p>gahhh im really nervous to see what i got =(</p>
<p>Are we talking about the same question here Michael209?</p>
<p>perfect- because of the intermediate value theorem you know that there’s a zero in between when the y value switches from being negative to positive</p>
<p>MathNerd-The graph came up from below the negative x-axis and turned twice and ended upwards on the positive x-axis. I really can’t explain it without an image, but I think you all should understand what I mean?</p>
<p>Michael- Thanks for the explanation!</p>
<p>oh… it was? i may have messed that up. :'(</p>
<p>What about the one with the sideways parabola, asking for the parametric equations. And it was a>o, because as x decreased the function approached -infinity, but as x increased the function approached infinity. So it would have to be a positive number to satisfy that.</p>
<p>No, they gave the graph of a -x^3 at the top. They drew out the curve.</p>
<p>^ yeah that’s how i remembered it! f(x) = ax^3, not -ax^3
like… hm
…^
…<em>…l
</em>l<strong>l</strong>l______
…l
l
v</p>
<p>(crossed from - to + first time, touched axis second time)</p>
<p>oh i thought the curve was ax^3?</p>
<p>Um. I’m confused. What is the correct answer finally for a>0,a<0? As it must be obvious, functions are not my strongpoint.</p>