Difference between revisions of "1954 AHSME Problems/Problem 42"

(Solution 1)
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==Solution 1==
 
==Solution 1==
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Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by <math>\frac{\frac{1}{2}}{2} = \frac{1}{4}</math>.
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Similarly, the x-coordinate of parabola 2 is given by <math>\frac{\frac{-1}{2}}{2} = -\frac{1}{4}</math>.
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From this information, we can deduce that <math>\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}</math>

Revision as of 23:26, 26 April 2020

Solution 1

Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by $\frac{\frac{1}{2}}{2} = \frac{1}{4}$.

Similarly, the x-coordinate of parabola 2 is given by $\frac{\frac{-1}{2}}{2} = -\frac{1}{4}$.

From this information, we can deduce that $\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}$