<p>A quadratic equation x^2 - 5x + 6 is graphed. Which of the following lines will cut this parabola at two points?</p>
<p>I) Y = 6
II) X = 2
III) Y = X + 1</p>
<p>A) I only
B) II only
C) III only
D) I and II
E) II and III</p>
<p>It’s a quadratic equation in Y= form – therefore, II can’t be correct because for each X value, there is only 1 Y value.</p>
<p>Next, you have to find the min/max of the equation. Because it’s positive, you’re looking for a minimum which occurs at -B/2A. For this equation, the minimum occurs at (5/1), or x= 5. The minimum of the equation is (5, 6).
Because of this, Y = 6 will only intersect the graph once, at the mimimum point.</p>
<p>That leaves Y = X+1 as the only right answer, so you’re looking for C.</p>
<p>ill tell you the basics. any quadratic equation is basically a function. according to the vertical lines rule, a line parallel to y axis can cut the function graph only once. in case of parabola a line parallel to x axis cuts the graph twice. so you do the math now. hope this helps.</p>
<p>@tizzy26; Hold on. This is a positive “X^2”, that means it will have a minimum point value, not a maximum point value. This means that the graph will look like V and not the other way around. </p>
<p>Or do I have it mixed up?</p>
<p>Waiting for more people to clarify.</p>
<p>I said minimum.</p>
<p>Exactly. A minimum point value means that the graph will be like this: V.
A maximum point value means the graph will look upside down (the opposite).</p>
<p>And this quadratic equation, indeed, has a minimum point value, not a maximum one, meaning that line Y = 6 does indeed cut it at the two points since it looks like “V”.</p>
<p>The minimum point value is at Y = 6, though. There is only 1 minimum, so it only intersects at 1 point.</p>
<p>well guys i think the person posting the original question might as well have gone to sleep listening to your chattering…lol</p>
<p>Thank you for your contribution. Let’s check out some more opinions.</p>
<p>well i wasnt being rude. just kidding. hope you dont mind.</p>
<p>Lol it’s fine. What exactly is your answer though, akashdip?</p>
<p>trololo. Just realized that I gave the right explanation and then did it wrong.</p>
<p>Again, the minimum is at -B / 2A. That puts the minimum at x = 5/2, not 5. </p>
<p>The minimum point is (2.5, -.25).</p>
<p>I and III are both correct, which means that none of the choices you gave are correct.</p>
<p>Lol, and that’s were the mind**** comes in. There was no choice for both I and III on the exam. It was either this or that. The choices given are exactly as they were on the exam.</p>
<p>well @tizzy26 did it right though. the coordinates are (-b/2a, (4ac-b^2)/4a). phew so long for such easy problem. the person posting this problem must be a freshman in high school. lol.</p>
<p>In that case, I’d ask the book writers :P</p>
<p>It wasn’t a book, lol. It was the real November 19th international exam, that I took yesterday.</p>
<p>in what “case” specifically?</p>
<p>OP is right. As transcribed from the exam the correct answer “I and III” is not offered as a choice.</p>
<p>SirWanksalot, is it possible that your recollection of one of the details may be wrong? A simple sign change – like y= -6 instead of y= 6 or something similar would lead to a proper list of choices. If not, and the problem is correctly transcribed, then the problem is clearly flawed and I expect that it would not be counted. It is not a particularly difficult problem and many test takers would have been equally surprised by the missing choice.</p>
<p>I am 100% sure, fogcity, that the question was written in this manner. The offered choices did not include both I and III. The offered choices were also “Y=6” and “Y=X+1”, and the quadratic equation was written in the same manner as presented. You may ask several other test takers for confirmation as well.</p>
<p>I guess under these circumstances, the question won’t be counted, huh?</p>
<p>Just curious: What # was this question?</p>