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

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== Problem 42==
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Consider the graphs of
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<cmath>(1)\qquad y=x^2-\frac{1}{2}x+2</cmath>
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and
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<cmath>(2)\qquad y=x^2+\frac{1}{2}x+2</cmath>
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on the same set of axis.
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These parabolas are exactly the same shape. Then:
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<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>
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==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>, since the parabolas are the same shape.

Latest revision as of 23:28, 26 April 2020

Problem 42

Consider the graphs of \[(1)\qquad y=x^2-\frac{1}{2}x+2\] and \[(2)\qquad y=x^2+\frac{1}{2}x+2\] on the same set of axis. These parabolas are exactly the same shape. Then:

$\textbf{(A)}\ \text{the graphs coincide.}\\ \textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).}\\ \textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).}\\ \textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}\\ \textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).}$


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 $\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).}$, since the parabolas are the same shape.

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