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.

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1955 AHSME Problems/Problem 24

Problem 24

Solution