<p>hey in the collegeboard blue book, can anyone explain how to do #16 on pg 859? thanks</p>
<p>Post the actual question - not all of us have the blue book.</p>
<p>i cant really, it has a graph of a parabola and this is the written portion</p>
<p>the figure above (like i said cant post it) shows the graph of a quadratic function h whose maximum calue is h(2). If h(a) = 0, which of the following could be the value of a?</p>
<p>-1
0
2
3
4</p>
<p>hope that helps, thanks</p>
<p>oh and does anyone know those equations using ax2 + bx + c, i think its something like maximum = b/c or something, yea i really dont know but would like to.</p>
<p>You cant solve this question without see the graph.
The value of "a" could be -1 or 5, the answer is choice A.
But its hard to explain.......I think I didn't solve this question in the best and easiest way...so I am also waiting for a good explaination.</p>
<h1>16. I can't help you too much without seeing the actual plot, but if it's a quadratic function with a maximum value, then the general function would be something like:</h1>
<p>h(x) = - b(x-2)^2 + c</p>
<p>where b is > 0 . At x=2, h(2) = c.</p>
<p>When h(a) =0, it just means the curve crosses the x-axis at x=0. This should happen at two possible values of a, both equally distant from x=2 i.e. a = 2 +/- k. a=0 or 4 looks like a good choice, since 2 +/- 2 gives us 0 or 4.</p>
<p>Quadratic function: f(x) = a x^2 + bx + c
The max/min for such functions (max if a < 0, min if a > 0) occurs when x = -b/2a; at this point, f( -b/2a) = -b^2/(4a) + c</p>
<p>phoson: yea i solved it by just kinda guessing and looking at it....not too reliable i think</p>
<p>Good explaination optimizerdad.....</p>
<p>I dont know if this is correct: I think when h(a) =0, the value of "a" could be the X value in the 2 crossing points on X-axis when Y=0.....
But you did't know the values of the "b" and "c".....
This problem has to be solved base on the graph.</p>
<p>By the way, the correct answer is A.</p>
<p>Vertical line through the vertex of a parabola is an axis of symmetry.
For our parabola an axis of symmetry is x_s=2.
Phson is correct: h(x)=0 in the 2 x-intercepts x1 and x2.
h(x1) = h(x2) = 0. Let's say x1 < x2.
Since h(a) = 0, there could be a=x1 or a=x2 only.
x1<0 on the diagram and there is only one negative value (-1) among answer choices, so a=-1 is the answer. (A).</p>
<hr>
<p>Additional (unnecessary) derivations:
x1 and x2 are equidistant from x<em>s=2. Since x1<0 on the diagram, the distance from x1 to x</em>s is greater the 2.
x2 must be more than 2 to the right of x_s=2,
x2 > 2+2,
x2 > 4.
Given values of "a" are all less or equal 4, so "a" can be x1 only.
x1<0 => a<0 => a=-1 is the only possible answer (A).</p>
<p>BTW: if x1=-1 then x2=5 - phson is right again.</p>
<p>Were the answer choices 1, 2, 3, 4, 5, the correct answer would be a=5.</p>
<p>Oops, typo in post#5. </p>
<p>'When h(a) =0, it just means the curve crosses the x-axis at x=0'
should read
'When h(a) =0, it just means the curve crosses the x-axis at x=a'</p>
<p>Sorry - I failed to give optimizerdad a credit: if he saw a diagram, he'd know that one of the h(x) roots is negative, and would derive a = 2 +/- 3, and a=-1 would be his choice.</p>
<p>Why can't the answer be 4? Isn't the second x-intercept 4 according to the graph? Why must it be negative? I hope someone sees this...</p>
<p>Ahh, I see. The neg intercept is greater than 2 away from the axis of symmetry, therefore it would have to be greater than two in the pos direction, so greater than 4. mm. I see this was already posted above as well. haha. My bad, there's my reading comprehension for you.</p>
<p>^daiea08 -- you must be done with most of the Consolidated List (or the whole Blue Book?) by now, cracking #16 on p.859.
Your math is good; CR needs a little bit of tweaking.:)</p>
<p>Incidentally, one student gave a very simple visual explanation to this question today.
If you squeeze that parabola towards its axis of symmetry x=2 so that its left branch passes through the origin, the left one will intercept the x-axis in x=4.
Now, if you expand the parabola back to its original form, the left x-intercept is <0, and the right one is >4.<br>
h(a)=0, a is the x-intercept.
a can't be B)0, C)2, D)3, E)4, that leaves A)-1.</p>
<p>^daiea08 -- you've corrupted the sacred Consolidated List thread!:eek: Well, just posted your question there... Still, start your own thread next time.
My response belongs to the SAT forum, but I don't want to start the thread for you. :p
So, 858/13.
Count how many unit squares are in the three pieces combined. The correct figure(s) should consist of the same number of unit squares.</p>