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

A parabola \(y = ax^{2}+bx + c\) has an axis of symmetry \(x =-\frac{b}{2a}\). For a parabola with vertex \((0,4)\) (equation \(y=ax^{2}+4\)) and another with vertex \((0, - 4)\) (equation \(y = mx^{2}-4\)). When we find the intersection points by setting \(ax^{2}+4=mx^{2}-4\), we get \(x^{2}=\frac{8}{m - a}\), \(x=\pm\sqrt{\frac{8}{m - a}}\).

Step2: Check the distance from the y - axis

The distance of a point \((x,y)\) from the \(y\) - axis is \(|x|\). For the intersection points \(x_{1}=\sqrt{\frac{8}{m - a}}\) and \(x_{2}=-\sqrt{\frac{8}{m - a}}\), the distance from the \(y\) - axis for \(x_{1}\) is \(|x_{1}|=\sqrt{\frac{8}{m - a}}\) and for \(x_{2}\) is \(|x_{2}|=\sqrt{\frac{8}{m - a}}\).

Answer:

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