Question

the vertex of a parabola that opens downward is at (0, 4). the vertex of a second parabola is at (0, -4). if the parabolas intersect at two points, which statement must be true?
the second parabola opens downward.
the second parabola opens upward.
the points of intersection are on the x - axis.
the points of intersection are of equal distance from the y - axis.

Explanation:

Step1: Analyze the symmetry of parabolas

Parabolas with vertices on the \(y -\)axis (\(x = 0\)) are symmetric about the \(y -\)axis. The equation of a parabola with vertex \((h,k)\) is \(y=a(x - h)^{2}+k\). For a parabola with vertex \((0,4)\), its equation is \(y = a_{1}x^{2}+4\) (since \(h = 0,k = 4\)) and it opens downward (\(a_{1}<0\)). For a parabola with vertex \((0,- 4)\), its equation is \(y=a_{2}x^{2}-4\) (\(h = 0,k=-4\)).

Step2: Use the property of intersection points

If \((x_{1},y_{1})\) is a point of intersection of the two parabolas \(y=a_{1}x^{2}+4\) and \(y=a_{2}x^{2}-4\), then when we substitute \(x=-x_{1}\) into the equations:
For \(y=a_{1}x^{2}+4\), \(y=a_{1}(-x_{1})^{2}+4=a_{1}x_{1}^{2}+4\)
For \(y=a_{2}x^{2}-4\), \(y=a_{2}(-x_{1})^{2}-4=a_{2}x_{1}^{2}-4\)
Since the two parabolas are symmetric about the \(y -\)axis (\(x = 0\)), if \((x,y)\) is a point of intersection, then \((-x,y)\) is also a point of intersection. The distance of a point \((x,y)\) from the \(y -\)axis is \(|x|\) and the distance of the point \((-x,y)\) from the \(y -\)axis is \(|-x|=|x|\)

Answer:

The points of intersection are of equal distance from the \(y -\)axis.