Difference between revisions of "1954 AHSME Problems/Problem 42"
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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 23:28, 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 
, since the parabolas are the same shape.
