Question
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?
Ask by Williams Cook. in South Africa
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The other x-intercept is at \(x = 5\).
Solution
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
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Extra Insights
To find another possible x-intercept, we can use the fact that for a quadratic function, the vertex forms the midpoint of the x-intercepts. Since one x-intercept is at x = 1, we can denote the other x-intercept as \( x_2 \). The x-coordinate of the vertex (3) can be expressed as the average of the x-intercepts: \[ 3 = \frac{1 + x_2}{2} \] Solving for \( x_2 \): \[ 6 = 1 + x_2 \\ x_2 = 5 \] Thus, the other x-intercept could be at \( x = 5 \).
