<p>PQRS is a square and points Q, R, and the orgin lie on the graph of y = ax^2, where a is a constant. If the area of the square is 64, what is the value of a?</p>
<p>The square is split right down the middle by the y axis... the P is to the left of the orgin by 4 points and the S is to the right of the orgin by 4 as well. Q is (-4, 8) and R (4, 8) R, S, and the orgin are on the parabola.. the orgin is the vertex of the parabola.... any help would be great... I thought the value of a would be 2 but it's wrong ,it should be 1/2... I just did 8-0/4-0..
like y2 - y1 / x2 - x1</p>
<p>With your question, a can be anything. It doesn't have to be 1/2.</p>
<p>It only works if P and S are on the x-axis.</p>
<p>The x-coordinates of Q and R are -4 and 4, respectively. Since P and S are on the x-axis, the height of the square must be 8. That means the y-coordinate of Q and the y-coordinate of R must be 8.</p>
<p>y = ax^2
8 = ? (4)^2
8 = ? x 16
? = 1/2.</p>
<p>Next time you have a question, please phrase it so that it is easier to understand. For example, "I just did 8-0/4-0.." Where did you get that? Why did you do 8-0, why did you do 4-0, and why did you divide them? Why did you think the value was 2? The equation y2-y1 / x2-x1 is used to find the slope of a straight line.</p>
<p>Area is 64 so each side is 8. 8/2 (the distance between each point is still 8, just goes from -4 to 4) is 4 so the points of the square are (4, 8), (4,0), (-4,0), (-4, 8). Use (4,8) and plug coordinates in y=ax^2. </p>
<p>No, because you don't know that the points of the square are (4,8), (4,0), (-4,0), and (-4,8). For example, you could have a square way up there but a would be so large that the parabola would be thin enough to touch Q and R.</p>
<p>The fact that the distance is 8 doesn't tell you anything about what the y-coordinates are. For example, the points could be (-10, 200), (-10,192), (10,192) and (10,200). That gives you 4 points of a square with side length 8. The points Q and R can be (-10,200) and (10,200). Then a would equal 2. So that possibility works as well. In other words, you don't have enough info to find the answer.</p>
<p>I'm sure that in the book there was a picture, where P and S were on the x-axis. </p>
<p>Yopua, you don't know that those four points you gave are actually the points. They don't have to be.</p>
<p>First of all, it is a square so distances have to be same and the only way you can have two points Q and R on the y=ax^2 is at the bottom since the y=0 only has 1 x-value.</p>
<p>"the only way you can have two points Q and R on the y=ax^2 is at the bottom since the y=0 only has 1 x-value."</p>
<p>No, for example, what if </p>
<p>the points could be (-10, 200), (-10,192), (10,192) and (10,200). That gives you 4 points of a square with side length 8. The points Q and R can be (-10,200) and (10,200). Then a would equal 2. So that possibility works as well. In other words, you don't have enough info to find the answer.</p>
<p>I don't understand your reasoning. Okay, when y = 0, there is only one x-value. That's true, but so what? What does that have to do with the square?</p>
<p>Again, -10 to 10 is not 8, it is 20. So what you have is a rectangle. I am saying y=0 so only 1 x-value because if you move the square up, you will have more than Q and R as points on ax^2, P and S would be too.</p>
<p>Oh, well, how about (-4, 32), (-4, 24), (4,24) and (4,32)? That makes a square.</p>
<p>Based on the original question, couldn't those points be P,Q,R,and S? Then a would be 2. Q and R are (-4,32) and (4,32). The area of the square is 64. So, why can't this be the answer?</p>
<p>Well that can't be it because all those four coordinates are in the first quadrant. Which would mean only 1 point would ever touch y=ax^2 since there x-int is 0 and y-int is 0. They said in the question atleast two points touch ax^2.</p>
<p>Yes, ax^2 does touch two quadrants. When you grapt y=2x^2, it does touch the points (-4,32) and (4,32). So if you let these be Q and R, would it work?</p>
<p>Okay you say the points are (-4,32), (-4,24), (4,32), (4,24).</p>
<p>Oh I see, I guess you have to assume only two of the points touch the ax^2 axis and those are Q and R and the others shouldn't to get the right answer.</p>
<p>Right. Yeah, when I help people with math I like to do it like this so that people can solve the problem for themselves rather than have me just tell them what the answer is.</p>