To determine the number of solutions to the linear-quadratic system of equations, we can analyze the graphs of the two equations provided.
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Equation 1 (Quadratic Equation): \[ y = -x^2 - 6x + 2 \] This is a downward-opening parabola (since the coefficient of \(x^2\) is negative).
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Equation 2 (Linear Equation): \[ y = -\frac{1}{11}x + 12 \] This is a straight line with a negative slope.
To find the points of intersection, we can set the two equations equal to each other: \[ -x^2 - 6x + 2 = -\frac{1}{11}x + 12 \] Multiplying both sides of the equation by 11 to eliminate the fraction: \[ -11x^2 - 66x + 22 = -x + 132 \] Rearranging terms gives: \[ -11x^2 - 65x - 110 = 0 \] Now we can use the discriminant to determine how many solutions exist, where the discriminant \(D\) is given by: \[ D = b^2 - 4ac \] For our quadratic:
- \(a = -11\)
- \(b = -65\)
- \(c = -110\)
Calculating the discriminant: \[ D = (-65)^2 - 4(-11)(-110) \] \[ D = 4225 - 4840 \] \[ D = -615 \]
Since the discriminant is negative, this implies that the quadratic equation has no real solutions, meaning that the graphs of the parabola and the line do not intersect.
Therefore, the correct response is: The graphs never intersect, so there is no solution.