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