Problem:
The parabola y=ax2+bx+c has vertex (p,p) and y-intercept (0,βp), where pξ =0. What is b ?
Answer Choices:
A. βp
B. 0
C. 2
D. 4
E. p
Solution:
A parabola with the given equation and with vertex (p,p) must have equation y=a(xβp)2+p. Because the y-intercept is (0,βp) and pξ =0, it follows that a=β2/p. Thus
y=βp2β(x2β2px+p2)+p=βp2βx2+4xβp
so b=4β.
The problems on this page are the property of the MAA's American Mathematics Competitions