Trying to follow step by step the graph of f(x) = square root of (1-2x^2). From my understanding the order of operations is supposed to be follwed when using graph transformations. If this is the case, I do not understand why the graph is undefined at y = 0. Algebraically, I understand. However, using graph tranformations, I do not understand. The tranformation seems to be out of order.
Graph Transformations Calculus
- Thread starter vnss
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If you square through, you see that the given function is the top half of the ellipse:
\(\displaystyle y^2=1-2x^2\)
\(\displaystyle 2x^2+y^2=1\)
\(\displaystyle \dfrac{x^2}{\left(\dfrac{1}{\sqrt{2}} \right)^2}+\dfrac{y^2}{1^2}=1\)
It is defined at \(\displaystyle y=0\) as the end-points \(\displaystyle \left(\pm\dfrac{1}{\sqrt{2}},0 \right)\) are on the curve.
\(\displaystyle y^2=1-2x^2\)
\(\displaystyle 2x^2+y^2=1\)
\(\displaystyle \dfrac{x^2}{\left(\dfrac{1}{\sqrt{2}} \right)^2}+\dfrac{y^2}{1^2}=1\)
It is defined at \(\displaystyle y=0\) as the end-points \(\displaystyle \left(\pm\dfrac{1}{\sqrt{2}},0 \right)\) are on the curve.
HallsofIvy
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It is not clear to me what you want. The "order of operations" is used when evaluating a formula. But what makes you think that "the graph is undefined when y= 0"? When \(\displaystyle x= 1/\sqrt{2}\), \(\displaystyle 2x^2= 2(1/2)= 1\) so that \(\displaystyle y= f(1/\sqrt{2})= \sqrt{1- 1}= 0\). The graph is defined at y= 0. Both \(\displaystyle (1/\sqrt{2}, 0)\) and \(\displaystyle (-1/\sqrt{2}, 0)\) are on the graph.
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It may appear undefined at y = 0 because the slope becomes undefined at y = 0, and this causes a problem with the way the calculator graphs. This problem would be avoided with a polar or parametric graph.