Which of the following functions will have the same x- intercept as its reflection about the origin

Question

y =x^2
y = 2x-5
y= -1/2x +6
y = (x-4)^2

Answer 1

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.

Answer 2

It can only be one

Answer 3

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!

Answer 4

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

Answer 5

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.