# 1955 AHSME Problems/Problem 24

## Problem 24

The function $4x^2-12x-1$: $\textbf{(A)}\ \text{always increases as }x\text{ increases}\\ \textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1}\\ \textbf{(C)}\ \text{cannot equal 0}\\ \textbf{(D)}\ \text{has a maximum value when }x\text{ is negative}\\ \textbf{(E)}\ \text{has a minimum value of-10}$

## Solution

We can use the process of elimination to narrow down the field substantially: $\textbf{(A)}\ \text{always increases as } x\text{ increases}$ is wrong due to the quadratic nature of the function. $\textbf{(B)}\ \text{always decreases as } x\text{ decreases to 1}$ is wrong due to the vertex being on the line $x = 1.5$. $f(x)$ would decrease all the way to $x = 1.5$, but increase from there. $\textbf{(C)}\ \text{cannot equal 0}$ is wrong due to the discriminant being greater than zero. $\textbf{(D)}\ \text{has a maximum value when}\ x \text{ is negative}$ is wrong due to the function not having a maximum value (the function opens up, so the vertex is the minimum value).

Therefore, $\textbf{(E)}$ is the correct answer (this can be verified by plugging in 1.5 (the x-coordinate of the vertex) in).

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 