To determine how many real solutions the equation \( f(x) = -2x^2 - 1 \) has, we need to analyze the function.
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The function is a quadratic function in standard form \( f(x) = ax^2 + bx + c \), where in this case \( a = -2 \), \( b = 0 \), and \( c = -1 \).
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The leading coefficient \( a = -2 \) is negative, indicating that the parabola opens downwards.
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The vertex form of a downward-opening parabola will either touch or cross the x-axis at most two times. To find the number of real solutions, we need to determine the vertex and check if it intersects the x-axis.
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To find the vertex (maximum point, since the parabola opens downwards), we can use the formula \( x = -\frac{b}{2a} \): \[ x = -\frac{0}{2(-2)} = 0. \] Now plug \( x = 0 \) back into \( f(x) \): \[ f(0) = -2(0)^2 - 1 = -1. \] So the vertex is at \( (0, -1) \).
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The value \( f(0) = -1 \) tells us that the highest point of the graph is below the x-axis, specifically at \( y = -1 \). This means the graph does not touch or cross the x-axis.
Based on this analysis, the answer to how many real solutions \( f(x) = -2x^2 - 1 \) has is:
no real solutions because the graphs neither touches nor crosses the x-axis.