Which of the following best characterizes the number of real solutions for the system of equations y = -2x+1 and y = -x^2 -1 (1 point)
The system of equations has two real solutions.
The system of equations has an infinite number of real solutions.
The system of equations has one real solution.
The system of equations has no real solutions
The system of equations has two real solutions.
To determine the number of real solutions for the system of equations, we need to find the points of intersection between the two equations.
Let's set the equations equal to each other:
-2x + 1 = -x^2 - 1
Now, let's rearrange the equation into quadratic form:
x^2 - 2x + 2 = 0
To find the number of real solutions, we can examine the discriminant of the quadratic equation: b^2 - 4ac.
In this case, a = 1, b = -2, and c = 2. Substituting these values into the discriminant formula:
(-2)^2 - 4(1)(2) = 4 - 8 = -4
Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the system of equations has no real solutions.
To determine the number of real solutions for the given system of equations, we need to find the intersection points of the two equations.
The first equation y = -2x + 1 represents a straight line with a slope of -2 and a y-intercept of 1.
The second equation y = -x^2 - 1 represents a downward opening parabola that crosses the y-axis at -1.
To find the intersection points, we need to set the two equations equal to each other:
-2x + 1 = -x^2 - 1
Rearranging this equation, we get:
x^2 - 2x + 2 = 0
Now, to determine the number of real solutions, we can use the discriminant (b^2 - 4ac) of the quadratic equation.
In this case, a = 1, b = -2, and c = 2.
The discriminant is given by (-2)^2 - 4(1)(2) = 4 - 8 = -4
Since the discriminant is negative (-4), there are no real solutions for the system of equations.
Therefore, the correct answer is:
The system of equations has no real solutions.