Problem:
Functions f and g are quadratic, g(x)=−f(100−x), and the graph of g contains the vertex of the graph of f. The four x-intercepts on the two graphs have x-coordinates x1,x2,x3, and x4, in increasing order, and x3−x2=150. The value of x4−x1 is m+np, where m,n, and p are positive integers, and p is not divisible by the square of any prime. What is m+n+p?
Answer Choices:
A. 602
B. 652
C. 702
D. 752
E. 802 Solution:
Let (h,k) be the vertex of the graph of f. Because the graph of f intersects the x-axis twice, we can assume that f(x)=a(x−h)2+k with a−k>0. Let s=a−k; then the x-intercepts of the graph of f are h±s. Because g(x)=−f(100−x)=−a(100−x−h)2−k, it follows that the x-intercepts of the graph of g are 100−h±s.
The graph of g contains the point (h,k); thus
k=f(h)=g(h)=−a(100−2h)2−k
from which h=50±22. Regardless of the sign in the expression for h, the four x-intercepts in order are
Because x3−x2=150, it follows that 150=s(2−2), that is s=150(1+22). Therefore x4−x1=s(2+2)=450+3002, and then m+n+p=450+300+2=752 .
OR
The graphs of f and g intersect the x-axis twice each. By symmetry, and because the graph of g contains the vertex of f, we can assume x1 and x3 are the roots\
of f, and x2 and x4 are the roots of g. A point (p,q) is on the graph of f if and only if (100−p,−q) is on the graph of g, so the two graphs are reflections of each other with respect to the point (50,0). Thus x2+x3=x1+x4=100, and since x3−x2=150, it follows that x2=−25 and x3=125. The average of x1 and x3=125 is h. It follows that x1=2h−125, from which x4=100−x1=225−2h, and x4−x1=350−4h.
Moreover, f(x)=a(x−x1)(x−x3)=a(x+125−2h)(x−125) and g(x)=−f(100−x)=−a(x+25)(x+2h−225). The vertex of the graph of f lies on the graph of g; thus
1=g(h)f(h)=−(h+25)(3h−225)(125−h)(h−125)
from which h=−25±752. However, h<x2<0; thus h=−25−752. Therefore x4−x1=450+3002 and then m+n+p=450+300+2=752.