Yes it is because 1/4 + 1/4 =2/4 and 2/4 equals 1/2
When I took the test, I selected "C. an arrangement in which you receive money now and pay it back later with fees" but I got it wrong. So I'm pretty sure the answer is "A. an arrangement in which you receive money, goods, or services now in exchange for the promise of payment later"
Stop with the links don’t work
<span><span>Graph <span>x2<span> = 4</span>y</span><span> and state the vertex, focus, axis of symmetry, and directrix.</span></span><span>This is the same graphing that I've done in the past: </span><span>y = (1/4)x2</span><span>. So I'll do the graph as usual:</span></span><span> </span><span>The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:<span>x2 = 4y</span> Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved<span>
(x – 0)2 = 4(y – 0)</span><span>This rearrangement "shows" that the vertex is at </span><span>(h, k) = (0, 0)</span><span>. The axis of symmetry is the vertical line right through the vertex: </span><span>x = 0</span>. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.<span>From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that </span><span>4p = 4</span><span>, so </span><span>p = 1</span><span>. Then the focus is one unit above the vertex, at </span>(0, 1)<span>, and the directrix is the horizontal line </span><span>y = –1</span>, one unit below the vertex.<span>vertex: </span>(0, 0)<span>; focus: </span>(0, 1)<span>; axis of symmetry: </span><span>x<span> = 0</span></span><span>; directrix: </span><span>y<span> = –1</span></span></span><span><span><span>Graph </span><span>y2<span> + 10</span>y<span> + </span>x<span> + 25 = 0</span></span>, and state the vertex, focus, axis of symmetry, and directrix.</span><span>Since the </span>y<span> is squared in this equation, rather than the </span>x<span>, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking </span>y<span>-values first and then finding the corresponding </span>x<span>-values for </span><span>x = –y2 – 10y – 25</span>:<span>To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:</span><span>y2 + 10y + 25 = –x</span> <span>
(y + 5)2 = –1(x – 0)</span><span>This tells me that </span><span>4p = –1</span><span>, so </span><span>p = –1/4</span><span>. Since the parabola opens to the left, then the focus is </span>1/4<span> units to the left of the vertex. I can see from the equation above that the vertex is at </span><span>(h, k) = (0, –5)</span><span>, so then the focus must be at </span>(–1/4, –5)<span>. The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is </span>1/4<span> units to the right of the vertex. Putting this all together, I get:</span><span>vertex: </span>(0, –5)<span>; focus: </span>(–1/4, –5)<span>; axis of symmetry: </span><span>y<span> = –5</span></span><span>; directrix: </span><span>x<span> = 1/4</span></span></span><span><span>Find the vertex and focus of </span><span>y2<span> + 6</span>y<span> + 12</span>x<span> – 15 = 0</span></span></span><span><span>The </span>y<span> part is squared, so this is a sideways parabola. I'll get the </span>y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.<span>y2 + 6y – 15 = –12x</span> <span><span>
y</span>2 + 6y + 9 – 15 = –12x + 9</span> <span>
(y + 3)2 – 15 = –12x + 9</span> <span>
(y + 3)2 = –12x + 9 + 15 = –12x + 24</span> <span>
(y + 3)2 = –12(x – 2)</span> <span>
(y – (–3))2 = 4(–3)(x – 2)</span></span><span><span>Then the vertex is at </span><span>(h, k) = (2, –3)</span><span> and the value of </span>p<span> is </span>–3<span>. Since </span>y<span> is squared and </span>p<span> is negative, then this is a sideways parabola that opens to the left. This puts the focus </span>3 units to the left of the vertex.<span>vertex: </span>(2, –3)<span>; focus: </span><span>(–1, –3)</span><span>
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