<p>In the xy-plane, the graph of the equation above assumes its maximum value at x = 2. What is the value of b?</p>
<pre><code>(A) –8
(B) –4
(C) 4
(D) 8
(E) 10
</code></pre>
<p>How do you do this? Plugging in every answer takes too long :D</p>
<p>EDIT: Please be very clear and explain every step thoroughly in solving this, as I have no clue how to go about it and would really like to be able to apply this info so that I can answer similar questions. Thanks!</p>
<p>Method 1:</p>
<p>If y=ax^2+bx+c, then the maximum occurs at x=-b/(2a).</p>
<p>So in this question we have -b/(2*-2) = 2, or b = 8.</p>
<p>Method 2 (using Calculus):</p>
<p>The derivative of y=-2x^2 + bx + 5 is y’ = -4x + b. We set the derivative to 0 and plug in x = 2 to get -4(2) + b = 0, or b = 8.</p>
<p>Plug x = 2 into the equation => y = -2x(2)^2 + 2b +5
<=> y = 2b - 3</p>
<p>Since y reaches its maximum value at x = 2, it’s easy to eliminate negative values because they make y negative. And we see that the larger b is, the larger y is. So b must be the largest positive value.</p>
<p>So my answer is (E).</p>
<p>But I’m not 100% sure since I suck at Math :)</p>
<p>I start to think I was wrong since DrSteve’s post reminds me of the x=-b/(2a) formula :</p>
<p>[The</a> Vertex of a Parabola](<a href=“http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html]The”>The Vertex of a Parabola)</p>
<p>This answers my question, thanks guys!</p>
<p>I like the way Dr Steve worked this out.</p>
<p>Thanks.</p>
<p>DrSteve’s Method 1 is by far the fastest - it takes less then 15 sec to get the answer.</p>
<p>It might happen though on the real test that you either forgot - or did not know to begin with - some formula needed to answer a question quickly.
If you don’t see a way around, come back to that question later (time allowing) and try it again, resorting, if necessary, to a no-brainer approach.</p>
<p>
</p>
<p>Enter in your TI graphing calculator ‘Y=’ editor
Y1 = -2x^2 -8x +5 - the vertex at x=-2 for (A) b=-8 </p>
<p>Y1 = -2x^2 +8x +5 - the vertex at x=2 for (D) b=8. Done in 30 sec (a bit longer if adjusting the Window settings).</p>
<p>=====================
A mistake here occurs when assuming that the vertex is in (2, 2b-3).
The fact that y=2b-3 when x=2 means only that a parabola passes through the point
(2, 2b-3).
For example, for b=-8 y=-2x^2 -8x +5, y(2)=-19, so the point (2,-19) is on the parabola, (but it is not the parabola’s vertex).
Plugging b=-8 into y=2b-3 will also get us y=-19 - the same point (2, -19).</p>
<p>++++++++++++++++++++
Now, a digression for the mathematically inclined.</p>
<p>If you graph parabolas y = -2x^2 + bx +5 for all given values of b
Y1 = -2x^2 -8x +5
Y2 = -2x^2 -4x +5
Y3 = -2x^2 +4x +5
Y4 = -2x^2 +8x +5
Y5 = -2x^2 +10x +5
you’ll see that all their vertices are on the parabola
Y6 = 2x^2 +5.
(In general, as b varies in y=ax^2 + bx + c, the vertex of this parabola traces parabola y=-ax^2+c.)</p>
<p>Here’s why.
y=-2x^2+bx+5,
the maximum occurs at x=-b/(2(-2)),
x=b/4.
The vertex y-coordinate
y=-2(b/4)^2 +b(b/4) +5
y=b^2/8 +5.</p>
<p>[You could also use the formula for y=ax^2+bx+c:
vertex y=-(b^2-4ac)/(4a).]</p>
<p>The vertex is in (x,y) = (b/4, b^2/8 +5).
Since x^2 = b^2/16
and y = b^2/8 +5, then
y = 2x^2 +5.
This means that the relation between the vertex coordinates is
y = 2x^2 +5.</p>
<p>Just thought that some people might get a kick out of it. :D</p>
