1998 AMC 12 Problem 14

Attempt Problem 14 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

A parabola has vertex at

(4,-5)

and has two

x

-intercepts, one positive and one negative. If this parabola is the graph of

y=ax^2+bx+c,

which of

a,

b,

and

c

must be positive?

only

a

only

b

only

c

a

and

b

only

none

The vertex lies below two intercepts, so determine the opening direction

Use the signs of the roots’ product and of

-\frac{b}{2a}=4

The parabola opens upward, so

a\gt0.

Its roots have opposite signs, so

\frac{c}{a}\lt0

and hence

c\lt0.

Also

-\frac{b}{2a}=4\gt0,

b\lt0.

Therefore only

a

must be positive, and A is correct.

Problem 14 in other years