Question 1210672: Precalculus
Chapter 1, section 1.2
Q. 41 (odd number question here).
The graph of an equation is given. The graph is a parabola opening to the left and right of the y-axis. The parabola opening to the left of the y-axis passes through the point (-1,0). The parabola opening to the right of the y-axis passes through the point (1,0).
A. Find the intercepts.
B. Indicate if the graph is symmetric with respect to the x-axis, y-axis, or the origin.
For A, I found the x-intercepts to be the points given on the graph. That is, the points (-1,0) and (1,0). Can I also represent the x-intercepts by saying x = -1 and x = 1? The two parabolic sketches do not cross the y-axis. So, my conclusion is that there are no y-intercepts. You say?
I don't know how to answer B.
Thank you
Answer by the_limit_DNE(3)   (Show Source):
You can put this solution on YOUR website! A. Yes, your solution path is correct so far. The only intercepts are the x-intercepts, (1, 0) and (-1, 0), which can be denotes as x=1 and x=-1 respectively.
B. Assuming that the points (1, 0) and (-1, 0) are vertices of the parabolas, then the graph has x-axis symmetry (x, -y). Furthermore, the graph would also have y-axis symmetry (-x, y) if this assumption is correct; by extension, the graph would also have origin-centered symmetry (-x, -y). The graph is only symmetric if the assumption is true.
C. The question mentioned a parabola "opening to the left and right of the x-axis", as well as a parabola "opening to the left" and "opening to the right"; the graph may actually be a hyperbola, not two parabolas.
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