<p>I will appreciate your help.</p>
<p>In the xy-coordinate plane, P and Q are different points that have the same y-coordinate and lie on the parabola whose equation is y= x^2 - 3x - 40. What is the x-coordinate of the midpoint of PQ?</p>
<p>(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2</p>
<p>The answer is D</p>
<p>Because of symmetry, the question is equivalent to asking: what is the x coordinate of the vertex? As has been discussed here recently (and often), that vertex can be found using x = -b/2a.</p>
<p>What has not been settled is whether this is something that the SAT actually ever tests…but it sure comes up here at CC a lot. And I guess it’s just one extra fact. Learn it if you’ll feel better :)</p>
<p>Here’s what I did. Since they have the same y coordinate u can arbitrarily pick one. I picked y=0 , and then I factored the equation. From that you get (x+5)(x-8) from that you get -5+8/2=3/2</p>
<p>Sent from my SCH-I535 using CC</p>
<p>@pckeller</p>
<p>I am a big supporter of having my more advanced students memorize this little fact. Certainly there has been no SAT problem where this is the only way to solve it, but I have seen several where this is the quickest way to solve it.</p>
<p>I just found where the first derivative equaled zero to get the answer.</p>
<p>Sent from my LG-E739 using CC</p>
<p>Of course you could just graph the thing on your calculator…</p>
<p>but for style points, I like speedread23’s solution best. I know the SAT does not award style points, but they should!</p>
<p>Another option to find the vertex would be to complete the square.</p>
<p>y= x² - 3x - 40
y = x² - 2(3/2)x + (3/2)² - (3/2)² - 40
y = (x - 3/2)² - (3/2)² - 40 </p>
<p>In the general equation of the parabola, y = a(x-h)² + k, (h,k) are the coordinates of the vertex. Here h = 3/2, which is the midpoint of any two points on the parabola that have the same y-coordinate.</p>
<p>The only drawback to speedread23’s method is in case you pick a value of y that does not give you real roots.</p>
<p>^But it is stil an interesting method – and you don’t ever actually need to figure out what y-value you have picked! Just pick something that allows you to factor in whatever way you happen to notice. Here’s what I mean:</p>
<p>Say the problem was the same question but with: y=x^2 + 12x + 4</p>
<p>If you set y = 0, you get nowhere.</p>
<p>But you could just say (x+10)(X+2) would be the factors of SOME properly chosen y-value. [In fact, it’s y = -16 but that info is not needed here.]</p>
<p>The roots would be -10 and -2. Their midpoint is at -6. Of course, that agrees with what you get if you use -b/2a. </p>
<p>I guess this is like a hybrid variant of completing the square. Not recommending this as the method I’d use on the SAT – just calling it interesting.</p>
<p>The -b/2a comes straight from the derivative – if y = ax^2 + bx + c = 0, then dy/dx = 2ax + b, setting to zero yields x = -b/2a.</p>
<p>However the SAT obviously doesn’t test calculus, so I’d only take derivatives if for some reason I forgot the -b/2a (e.g. if I thought it was b/2a or -b/a).</p>
<p>@pckeller I agree it is an interesting approach, and I like how you suggested to factor it in any possible way. That is a lot more flexible than picking a specific value that could cause confusion if one ends up with no real roots. </p>
<p>All this discussion made me think of another way:</p>
<p>y = x² - 3x - 40 </p>
<p>Find the y-intercept of the parabola when x is zero, replace x = 0 in the above quadratic.
This gives us: </p>
<p>y = 0 - 0 - 40 = -40 </p>
<p>then replace y with -40 in the above quadratic equation, and find the corresponding x-value:</p>
<p>y= -40 = x² - 3x - 40 and solve for x:
0 = x² - 3x or x(x-3) =0, one solution is 0 which we have already used, the other one is x=3. Now find the point halfway between 0 and 3, which is again 3/2 and is the x-coordinate of the vertex of the parabola.</p>