Difference between revisions of "1954 AHSME Problems/Problem 42"
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+ | == Problem 42== | ||
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+ | Consider the graphs of | ||
+ | <cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath> | ||
+ | and | ||
+ | <cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath> | ||
+ | on the same set of axis. | ||
+ | These parabolas are exactly the same shape. Then: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{the graphs coincide.}\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).} </math> | ||
+ | |||
+ | |||
==Solution 1== | ==Solution 1== | ||
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>. | 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>. | Similarly, the x-coordinate of parabola 2 is given by <math>\frac{\frac{-1}{2}}{2} = -\frac{1}{4}</math>. | ||
− | 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> | + | 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>, since the parabolas are the same shape. |
Latest revision as of 22:28, 26 April 2020
Problem 42
Consider the graphs of and on the same set of axis. These parabolas are exactly the same shape. Then:
Solution 1
Let us consider the vertices of the two parabolas. The x-coordinate of parabola 1 is given by .
Similarly, the x-coordinate of parabola 2 is given by .
From this information, we can deduce that , since the parabolas are the same shape.