Question
Which of the following functions will have the same x- intercept as its reflection about the origin
y =x^2
y = 2x-5
y= -1/2x +6
y = (x-4)^2
y =x^2
y = 2x-5
y= -1/2x +6
y = (x-4)^2
Answers
Answer
It can only be one
Answered by
oobleck
still wrong. Unless the x-intercept is at (0,0), reflecting about the origin will produce a different intercept.
The point (x,y) becomes (-x,-y) so (5/2,0) becomes (-5/2,0) which is the x-intercept of y=2x+5
The point (x,y) becomes (-x,-y) so (5/2,0) becomes (-5/2,0) which is the x-intercept of y=2x+5
Answered by
GPT-4o mini
To determine which functions will have the same x-intercept as their reflection about the origin, we first need to find the x-intercepts of each function and then see if they remain the same after reflection.
The x-intercept of a function \( f(x) \) occurs where \( f(x) = 0 \).
1. **Function: \( y = x^2 \)**
Set \( y = 0 \):
\( x^2 = 0 \)
\( x = 0 \)
**X-intercept**: \( (0, 0) \)
The reflection about the origin is \( y = -x^2 \).
Set \( -x^2 = 0 \):
\( x^2 = 0 \)
**Same x-intercept**: \( (0, 0) \)
2. **Function: \( y = 2x - 5 \)**
Set \( y = 0 \):
\( 2x - 5 = 0 \)
\( 2x = 5 \)
\( x = \frac{5}{2} \)
**X-intercept**: \( \left( \frac{5}{2}, 0 \right) \)
The reflection about the origin is \( y = -2x + 5 \).
Set \( -2x + 5 = 0 \):
\( 2x = 5 \)
\( x = \frac{5}{2} \)
**Same x-intercept**: \( \left( \frac{5}{2}, 0 \right) \)
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
Set \( y = 0 \):
\( -\frac{1}{2}x + 6 = 0 \)
\( \frac{1}{2}x = 6 \)
\( x = 12 \)
**X-intercept**: \( (12, 0) \)
The reflection about the origin is \( y = \frac{1}{2}x - 6 \).
Set \( \frac{1}{2}x - 6 = 0 \):
\( \frac{1}{2}x = 6 \)
\( x = 12 \)
**Same x-intercept**: \( (12, 0) \)
4. **Function: \( y = (x-4)^2 \)**
Set \( y = 0 \):
\( (x-4)^2 = 0 \)
\( x - 4 = 0 \)
\( x = 4 \)
**X-intercept**: \( (4, 0) \)
The reflection about the origin is \( y = -(x-4)^2 \).
Set \( -(x-4)^2 = 0 \):
\( (x-4)^2 = 0 \)
\( x - 4 = 0 \)
\( x = 4 \)
**Same x-intercept**: \( (4, 0) \)
### Conclusion:
All four functions have the same x-intercept as their reflections about the origin:
- \( y = x^2 \): x-intercept at \( (0, 0) \)
- \( y = 2x - 5 \): x-intercept at \( \left( \frac{5}{2}, 0 \right) \)
- \( y = -\frac{1}{2}x + 6 \): x-intercept at \( (12, 0) \)
- \( y = (x-4)^2 \): x-intercept at \( (4, 0) \)
Thus, all functions listed have the same x-intercept as their reflections about the origin.
The x-intercept of a function \( f(x) \) occurs where \( f(x) = 0 \).
1. **Function: \( y = x^2 \)**
Set \( y = 0 \):
\( x^2 = 0 \)
\( x = 0 \)
**X-intercept**: \( (0, 0) \)
The reflection about the origin is \( y = -x^2 \).
Set \( -x^2 = 0 \):
\( x^2 = 0 \)
**Same x-intercept**: \( (0, 0) \)
2. **Function: \( y = 2x - 5 \)**
Set \( y = 0 \):
\( 2x - 5 = 0 \)
\( 2x = 5 \)
\( x = \frac{5}{2} \)
**X-intercept**: \( \left( \frac{5}{2}, 0 \right) \)
The reflection about the origin is \( y = -2x + 5 \).
Set \( -2x + 5 = 0 \):
\( 2x = 5 \)
\( x = \frac{5}{2} \)
**Same x-intercept**: \( \left( \frac{5}{2}, 0 \right) \)
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
Set \( y = 0 \):
\( -\frac{1}{2}x + 6 = 0 \)
\( \frac{1}{2}x = 6 \)
\( x = 12 \)
**X-intercept**: \( (12, 0) \)
The reflection about the origin is \( y = \frac{1}{2}x - 6 \).
Set \( \frac{1}{2}x - 6 = 0 \):
\( \frac{1}{2}x = 6 \)
\( x = 12 \)
**Same x-intercept**: \( (12, 0) \)
4. **Function: \( y = (x-4)^2 \)**
Set \( y = 0 \):
\( (x-4)^2 = 0 \)
\( x - 4 = 0 \)
\( x = 4 \)
**X-intercept**: \( (4, 0) \)
The reflection about the origin is \( y = -(x-4)^2 \).
Set \( -(x-4)^2 = 0 \):
\( (x-4)^2 = 0 \)
\( x - 4 = 0 \)
\( x = 4 \)
**Same x-intercept**: \( (4, 0) \)
### Conclusion:
All four functions have the same x-intercept as their reflections about the origin:
- \( y = x^2 \): x-intercept at \( (0, 0) \)
- \( y = 2x - 5 \): x-intercept at \( \left( \frac{5}{2}, 0 \right) \)
- \( y = -\frac{1}{2}x + 6 \): x-intercept at \( (12, 0) \)
- \( y = (x-4)^2 \): x-intercept at \( (4, 0) \)
Thus, all functions listed have the same x-intercept as their reflections about the origin.
Answered by
GPT-4o mini
Thank you for your patience! Let's clarify the solution step by step to identify which function has the same x-intercept as its reflection about the origin.
1. **Function: \( y = x^2 \)**
- X-intercept:
\[
x^2 = 0 \implies x = 0 \implies (0, 0)
\]
- Reflection about the origin: \( y = -x^2 \)
- X-intercept:
\[
-x^2 = 0 \implies x = 0 \implies (0, 0)
\]
2. **Function: \( y = 2x - 5 \)**
- X-intercept:
\[
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
- Reflection about the origin: \( y = -2x + 5 \)
- X-intercept:
\[
-2x + 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
- X-intercept:
\[
-\frac{1}{2}x + 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
- Reflection about the origin: \( y = \frac{1}{2}x - 6 \)
- X-intercept:
\[
\frac{1}{2}x - 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
4. **Function: \( y = (x-4)^2 \)**
- X-intercept:
\[
(x-4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \implies (4, 0)
\]
- Reflection about the origin: \( y = -(x-4)^2 \)
- X-intercept:
\[
-(x-4)^2 = 0 \implies (x-4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \implies (4, 0)
\]
### Final Result:
- All functions we checked also maintain their x-intercepts under reflection, but to identify just **one** function having the same x-intercept and confirm the x-intercepts:
- Any of the functions can be said to have the same x-intercept as their reflection:
- One clear option is \( y = 2x - 5 \) and its reflection \( y = -2x + 5 \) both have the x-intercept of \( \left( \frac{5}{2}, 0 \right) \).
But since we need to specify just **one**, we could present \( y = 2x - 5 \).
Please clarify if you need a definite answer based on specific selection criteria!
1. **Function: \( y = x^2 \)**
- X-intercept:
\[
x^2 = 0 \implies x = 0 \implies (0, 0)
\]
- Reflection about the origin: \( y = -x^2 \)
- X-intercept:
\[
-x^2 = 0 \implies x = 0 \implies (0, 0)
\]
2. **Function: \( y = 2x - 5 \)**
- X-intercept:
\[
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
- Reflection about the origin: \( y = -2x + 5 \)
- X-intercept:
\[
-2x + 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
- X-intercept:
\[
-\frac{1}{2}x + 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
- Reflection about the origin: \( y = \frac{1}{2}x - 6 \)
- X-intercept:
\[
\frac{1}{2}x - 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
4. **Function: \( y = (x-4)^2 \)**
- X-intercept:
\[
(x-4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \implies (4, 0)
\]
- Reflection about the origin: \( y = -(x-4)^2 \)
- X-intercept:
\[
-(x-4)^2 = 0 \implies (x-4)^2 = 0 \implies x - 4 = 0 \implies x = 4 \implies (4, 0)
\]
### Final Result:
- All functions we checked also maintain their x-intercepts under reflection, but to identify just **one** function having the same x-intercept and confirm the x-intercepts:
- Any of the functions can be said to have the same x-intercept as their reflection:
- One clear option is \( y = 2x - 5 \) and its reflection \( y = -2x + 5 \) both have the x-intercept of \( \left( \frac{5}{2}, 0 \right) \).
But since we need to specify just **one**, we could present \( y = 2x - 5 \).
Please clarify if you need a definite answer based on specific selection criteria!
Answered by
GPT-4o mini
You are correct, and I apologize for the misunderstanding in my earlier responses. To clarify, reflecting a point about the origin indeed transforms the point \((x,y)\) to \((-x,-y)\). This means that the x-intercept after reflection will be at \((-x, 0)\).
Let's re-evaluate the x-intercepts of each function and check which ones will produce the same x-intercept upon reflection about the origin:
1. **Function: \( y = x^2 \)**
- **X-intercept**:
\[
x^2 = 0 \implies x = 0 \implies (0, 0)
\]
- **Reflection**: remains \((0, 0)\) after reflection
2. **Function: \( y = 2x - 5 \)**
- **X-intercept**:
\[
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
- **Reflection** leads to x-intercept at \(-\frac{5}{2}\)
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
- **X-intercept**:
\[
-\frac{1}{2}x + 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
- **Reflection** leads to x-intercept at \(-12\)
4. **Function: \( y = (x - 4)^2 \)**
- **X-intercept**:
\[
(x - 4)^2 = 0 \implies x = 4 \implies (4, 0)
\]
- **Reflection** leads to x-intercept at \(-4\)
### Conclusion:
In evaluating all functions, only the function \( y = x^2 \) maintains the same x-intercept \((0, 0)\) when reflected across the origin:
- \( y = 2x - 5 \) reflects to have an x-intercept at \(-\frac{5}{2}\).
- \( y = -\frac{1}{2}x + 6 \) reflects to have an x-intercept at \(-12\).
- \( y = (x - 4)^2 \) reflects to have an x-intercept at \(-4\).
Thus, the **only function** that has the same x-intercept at the origin \((0, 0)\) after reflection about the origin is **\( y = x^2 \)**.
Thank you for your patience, and I appreciate your clarification.
Let's re-evaluate the x-intercepts of each function and check which ones will produce the same x-intercept upon reflection about the origin:
1. **Function: \( y = x^2 \)**
- **X-intercept**:
\[
x^2 = 0 \implies x = 0 \implies (0, 0)
\]
- **Reflection**: remains \((0, 0)\) after reflection
2. **Function: \( y = 2x - 5 \)**
- **X-intercept**:
\[
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} \implies \left( \frac{5}{2}, 0 \right)
\]
- **Reflection** leads to x-intercept at \(-\frac{5}{2}\)
3. **Function: \( y = -\frac{1}{2}x + 6 \)**
- **X-intercept**:
\[
-\frac{1}{2}x + 6 = 0 \implies \frac{1}{2}x = 6 \implies x = 12 \implies (12, 0)
\]
- **Reflection** leads to x-intercept at \(-12\)
4. **Function: \( y = (x - 4)^2 \)**
- **X-intercept**:
\[
(x - 4)^2 = 0 \implies x = 4 \implies (4, 0)
\]
- **Reflection** leads to x-intercept at \(-4\)
### Conclusion:
In evaluating all functions, only the function \( y = x^2 \) maintains the same x-intercept \((0, 0)\) when reflected across the origin:
- \( y = 2x - 5 \) reflects to have an x-intercept at \(-\frac{5}{2}\).
- \( y = -\frac{1}{2}x + 6 \) reflects to have an x-intercept at \(-12\).
- \( y = (x - 4)^2 \) reflects to have an x-intercept at \(-4\).
Thus, the **only function** that has the same x-intercept at the origin \((0, 0)\) after reflection about the origin is **\( y = x^2 \)**.
Thank you for your patience, and I appreciate your clarification.
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