Question

Which function shares no real solutions with the quadratic function in the graph?

(1 point)
Responses

y=−12x+6
y equals Start Fraction 1 over 2 End Fraction x plus 6

y=12x−1
y equals Start Fraction 1 over 2 End Fraction x minus 1

y=x+1
y equals x plus 1

y=x2

Answer 1

To determine which function shares no real solutions with the quadratic function \( y = x^2 \), we need to analyze the possible intersections between \( y = x^2 \) and the linear functions provided. A lack of real solutions means that the linear function must not intersect the parabola at any point.

1. **For \( y = -12x + 6 \)**:
Set \( x^2 = -12x + 6 \).
Rearranging gives: \( x^2 + 12x - 6 = 0 \).
This is a quadratic equation that will have real solutions depending on the discriminant \( b^2 - 4ac = 12^2 - 4(1)(-6) = 144 + 24 = 168 \), which is positive, indicating 2 real solutions.

2. **For \( y = \frac{1}{2}x + 6 \)**:
Set \( x^2 = \frac{1}{2}x + 6 \).
Rearranging gives: \( x^2 - \frac{1}{2}x - 6 = 0 \).
The discriminant here is \( (-\frac{1}{2})^2 - 4(1)(-6) = \frac{1}{4} + 24 = \frac{97}{4} \), which is positive, indicating 2 real solutions.

3. **For \( y = 12x - 1 \)**:
Set \( x^2 = 12x - 1 \).
Rearranging gives: \( x^2 - 12x + 1 = 0 \).
The discriminant is \( (-12)^2 - 4(1)(1) = 144 - 4 = 140 \), which is positive, indicating 2 real solutions.

4. **For \( y = \frac{1}{2} x - 1 \)**:
Set \( x^2 = \frac{1}{2}x - 1 \).
Rearranging gives: \( x^2 - \frac{1}{2}x + 1 = 0 \).
The discriminant is \( (-\frac{1}{2})^2 - 4(1)(1) = \frac{1}{4} - 4 = \frac{1 - 16}{4} = -\frac{15}{4} \), which is negative, indicating no real solutions.

5. **For \( y = x + 1 \)**:
Set \( x^2 = x + 1 \).
Rearranging gives: \( x^2 - x - 1 = 0 \).
The discriminant is \( (-1)^2 - 4(1)(-1) = 1 + 4 = 5 \), which is positive, indicating 2 real solutions.

Based on this analysis, the function that shares no real solutions with the quadratic function \( y = x^2 \) is:

**\( y = \frac{1}{2}x - 1 \)**.