Question

Which of the following functions will have the same x-intercept(s) as its reflection about the origin?
y = 2x - 5
y = x^2
y = (x - 4)^2
y = -1/2x + 6

Answer 1

To determine which of the given functions will have the same x-intercept(s) as its reflection about the origin, we need to first understand what the reflection of a function about the origin means.

When a function \( y = f(x) \) is reflected about the origin, it transforms into \( y = -f(-x) \). The x-intercepts of a function are the points where \( y = 0 \). Therefore, we need to find the x-intercepts of both the original function and its reflection, and see if they are the same.

Let's analyze each function one by one.

1. **Function: \( y = 2x - 5 \)**
- To find the x-intercept, set \( y = 0 \):
\[
0 = 2x - 5 \implies 2x = 5 \implies x = \frac{5}{2}
\]
- Reflecting the function yields \( y = -(2(-x) - 5) = -(-2x - 5) = 2x + 5 \).
- To find the x-intercept of the reflection:
\[
0 = 2x + 5 \implies 2x = -5 \implies x = -\frac{5}{2}
\]
- The x-intercepts are different.

2. **Function: \( y = x^2 \)**
- Finding the x-intercept:
\[
0 = x^2 \implies x = 0
\]
- Reflecting the function yields \( y = -(-x)^2 = -x^2 \).
- The x-intercept of the reflection:
\[
0 = -x^2 \implies x^2 = 0 \implies x = 0
\]
- The x-intercepts are the same.

3. **Function: \( y = (x - 4)^2 \)**
- Finding the x-intercept:
\[
0 = (x - 4)^2 \implies x - 4 = 0 \implies x = 4
\]
- Reflecting the function yields \( y = -((-x) - 4)^2 = -(-x - 4)^2 \).
- The x-intercept of the reflection:
\[
0 = -(x + 4)^2 \implies (x + 4)^2 = 0 \implies x = -4
\]
- The x-intercepts are different.

4. **Function: \( y = -\frac{1}{2}x + 6 \)**
- Finding the x-intercept:
\[
0 = -\frac{1}{2}x + 6 \implies \frac{1}{2}x = 6 \implies x = 12
\]
- Reflecting the function yields \( y = -\left(-\frac{1}{2}(-x) + 6\right) = -\left(\frac{1}{2}x + 6\right) = -\frac{1}{2}x - 6 \).
- The x-intercept of the reflection:
\[
0 = -\frac{1}{2}x - 6 \implies -\frac{1}{2}x = 6 \implies x = -12
\]
- The x-intercepts are different.

In summary, the only function that maintains the same x-intercept as its reflection about the origin is:

**\( y = x^2 \)**.