<p>The graph of y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2 lies above the line y = bx + c except at three values of , where the graph and the line intersect. What is the largest of those values? </p>
<p>A. 4
B. 5
C. 6
D. 7
E. 8</p>
<p>Please explain how you got the answer. Thanks!</p>
<p>I dont see how the line can intercept the graph at 3 points, I used a graphing program to graph the rational function and it appears the most times that a line can intercept it is 2.</p>
<p>This is as far as I got:</p>
<p>Obviously it is tangent at three points, thus:</p>
<p>x^6 - 10x^5 + 29x^4 - 4x^3 + ax^ - bx - c = 0
x^6 - 10x^5 + 29x^4 - 4x^3 + ax^ - bx - c = [(x-p)(x-q)(x-r)]^2</p>
<p>(Its squared because they are double roots. Also, “p,q,r” are the roots.)</p>
<p>This is where I got stuck…</p>
<p>Huh, a polynomial with the largest power of even looks almost parabolic. I can’t see that it’s feasible for a line to intersect this curve 3 times.</p>
<p>-.-</p>
<p>6 degree polynomial means 5 local maxima/minima => 3 upper tangents (dy/dx = 0) 2 lower tangents. Its definitely feasible.</p>
<p>a 6degree polynomial doesnt always have 5 maxima/minima, this one appears to have 1.</p>
<p>Umm, what are you talking about? I wouldn’t get hung up on that detail, I just need help applying Vieta’s…</p>
<p>Reading the question, it said lies above at all except those points, where it intersects. Since the second function is a line, and the larger polynomial must always be above or equal to the second function, the intersection points will all be local minimums of the polynomial.
At 3 values of _? Is there more to the question, like at only 3 certain values of a b and c? Or is it meaning at 3 values its not the first function isn’t greater than the second, because they’re equal? If its the second, I’d just plug in some numbers for a b and c and see how the graph and minimums look.</p>
<p>Also, being a polynomial to the 6th power, you can factor out x^2. This results in a 0 at x=0, so you know it passes through the origin.</p>
<p>Not sure what else you can do from there, to find the answer, as I do not own a graphing calculator heheh.</p>
<p>BTW this is from a no calculator test!</p>
<p>Oh wow thanks factoring out the x^2 is useful</p>
<p>x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2 - bx - c = roots^2
x^2 ( x^4 - 10x^3 + 29x^2 - 4x + a ) - bx - c = [(x-p)(x-r)(x-q)]^2</p>
<p>…eek</p>
<p>The answer is A) 4.</p>
<p>My question: where did you get this problem, because I doubt you got this from SAT? lol</p>
<p>Here is my solution (sorry in advance if my wording is too confusing):
Let the roots be a, b, c. Given x^6 - 10x^5 + 29x^4 - 4x^3… and that our function is of the form (x-a)^2(x-b)^2(x-c)^2, by Vieta, 10 = 2(a+b+c), or 5 = a+b+c, and 29 = a^2+b^2+c^2 + 4(ab+bc+ca). The former is the sum of all the roots, and the latter is the sum of all the products of two roots. Similarly, 4 = the sum of all the products of three roots. </p>
<p>Now we will proceed to do some manipulations.
(a+b+c)^2 = 25 = a^2+b^2+c^2 + 2(ab+bc+ca)
Subtract that from 29 = a^2+b^2+c^2 + 4(ab+bc+ca) to obtain 4 = 2(ab+bc+ca), or 2 = ab+bc+ca.
Subtract 2= ab+bc+ca from 25 = a^2+b^2+c^2 + 2(ab+bc+ca) to get 23=a^2+b^2+c^2 +ab+bc+ca
Notice that a^2+b^2+c^2 +ab+bc+ca = ((a+b)^2 + (b+c)^2 + (c+a)^2)/2
So, 46 = (a+b)^2 + (b+c)^2 + (c+a)^2
Now, I’m going to be sneaky and try some values of a+b, b+c, c+a to make that and 5=a+b+c, or 10 = (a+b) + (b+c) + (c+a) work out. I get values that work: a+b = 6, b+c = 1, c+a=3. BTW all of these are done without the loss of generality of a,b, and c. The values we get are of course non rigorous, but we will see later that they work out. </p>
<p>With a+b = 6, b+c = 1, c+a=3, we do subtractions between them to obtain a-c = 5, a-b=2, b-c=3. Add a-c to c+a, a-b to a+b, b-c to b+c, to obtain 2a=8, 2b=4, 2c=-2, or a=4, b=2, c=-1. </p>
<p>Now comes the verification using “4 = the sum of all the products of three roots.” We know that if this works out, then we have obtained a unique solution, since we have three variables and used three equations. So there are 20 terms (6c3) in this sum of all products of three roots. Using 4, 4, 2, 2, -1, -1, I calculated this out (the (tedious) process of which I will omit here), and voila, obtained 4, which is what we were looking for. </p>
<p>NOW,
Although the solution looks long (i tried to make it detailed) this didn’t take as long as it might seem (less than 5 minutes probably).
However, there is probably a much much simpler way to do this, which I hope someone will be able to come up with!</p>