Difference between revisions of "1954 AHSME Problems/Problem 42"
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| + | == Problem 42== | ||
| + | |||
| + | 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> | ||
| + | |||
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==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>. | ||
Revision as of 22:28, 26 April 2020
Problem 42
Consider the graphs of
![\[(1)\qquad y=x^2-\frac{1}{2}x+2\]](http://latex.artofproblemsolving.com/8/4/d/84deece22a8eb46f2ccd71a02b810b0bb22c4f32.png)
and
![\[(2)\qquad y=x^2+\frac{1}{2}x+2\]](http://latex.artofproblemsolving.com/e/f/2/ef280510c51f3df7b80e29c0cc506f4ec9a192ee.png)
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 
