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