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 consider the reflection about the origin mathematically.

The reflection of a point \((x, y)\) about the origin is given by \((-x, -y)\). Therefore, for a function \(y = f(x)\), its reflection will be described by the function \(y = -f(-x)\).

Next, we need to find the x-intercepts of the original functions and their respective reflections:

1. **Function: \(y = 2x - 5\)**

- Find x-intercept by setting \(y = 0\):
\[
0 = 2x - 5 \implies 2x = 5 \implies x = \frac{5}{2}
\]

- Reflection: \(y = -f(-x) = - (2(-x) - 5) = -(-2x - 5) = 2x + 5\)
- Find x-intercept:
\[
0 = 2x + 5 \implies 2x = -5 \implies x = -\frac{5}{2}
\]

- Conclusion: x-intercepts do not match.

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

- Find x-intercept by setting \(y = 0\):
\[
0 = x^2 \implies x = 0
\]

- Reflection: \(y = -f(-x) = -((-x)^2) = -x^2\)
- Find x-intercept:
\[
0 = -x^2 \implies x = 0
\]

- Conclusion: x-intercepts match.

3. **Function: \(y = (x - 4)^2\)**

- Find x-intercept by setting \(y = 0\):
\[
0 = (x - 4)^2 \implies x - 4 = 0 \implies x = 4
\]

- Reflection: \(y = -f(-x) = -(-x - 4)^2 = -(x + 4)^2\)
- Find x-intercept:
\[
0 = -(x + 4)^2 \implies x + 4 = 0 \implies x = -4
\]

- Conclusion: x-intercepts do not match.

4. **Function: \(y = -\frac{1}{2}x + 6\)**

- Find x-intercept by setting \(y = 0\):
\[
0 = -\frac{1}{2}x + 6 \implies \frac{1}{2}x = 6 \implies x = 12
\]

- Reflection: \(y = -f(-x) = -\left( -\frac{1}{2}(-x) + 6 \right) = -\left( \frac{1}{2}x + 6 \right) = -\frac{1}{2}x - 6\)
- Find x-intercept:
\[
0 = -\frac{1}{2}x - 6 \implies -\frac{1}{2}x = 6 \implies x = -12
\]

- Conclusion: x-intercepts do not match.

Based on this analysis, the only function that has the same x-intercept as its reflection about the origin is:

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