A quadratic function has its vertex at (3, -2) and opens upwards. If one of its x-intercepts is at x = 1, what could be another possible x-intercept?

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To find the other x-intercept of the quadratic function, let's use the properties of the vertex and symmetry of a parabola.

1. **Vertex Form of a Quadratic Function:**
   The vertex form is given by:
   $$
   y = a(x - h)^2 + k
   $$
   where $$(h, k)$$ is the vertex. Given the vertex $$(3, -2)$$, the equation becomes:
   $$
   y = a(x - 3)^2 - 2
   $$
   Since the parabola opens upwards, $$a > 0$$.

2. **Symmetry of the Parabola:**
   The axis of symmetry is the vertical line $$x = 3$$. If one x-intercept is at $$x = 1$$, the other x-intercept must be symmetrically located on the other side of the axis of symmetry.

3. **Finding the Other X-Intercept:**
   - The distance from the x-intercept $$x = 1$$ to the axis of symmetry $$x = 3$$ is $$2$$ units to the left.
   - Therefore, the other x-intercept must be $$2$$ units to the right of $$x = 3$$, which is $$x = 5$$.

**Thus, the other possible x-intercept is at $$x = 5$$.**

Answer: 5
The other x-intercept is at $$x = 5$$.

A quadratic function has its vertex at (3, -2) and opens upwards. If one of its x-intercepts is at x = 1, what could be another possible x-intercept?

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