To determine how many real solutions the function \( f(x) = -2x^2 - 1 \) has, we can analyze the characteristics of the function, specifically its graph.
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Function Type: This is a quadratic function in the form \( f(x) = ax^2 + bx + c \) where \( a = -2 \), \( b = 0 \), and \( c = -1 \).
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Direction of Opening: Since \( a < 0 \), the parabola opens downwards.
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Vertex: The vertex of the parabola is located at \( x = -\frac{b}{2a} = -\frac{0}{2(-2)} = 0 \). Plugging \( x = 0 \) into the function, we find the vertex:
\[ f(0) = -2(0)^2 - 1 = -1 \]
This means the vertex is at the point (0, -1).
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Y-Intercept: The value of \( f(0) \) also gives us the y-intercept, which is \( -1 \). The graph intersects the y-axis at this point.
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X-Intercepts: To find the x-intercepts (real solutions), we need to set \( f(x) = 0 \):
\[ -2x^2 - 1 = 0 \]
Rearranging gives:
\[ -2x^2 = 1 \implies x^2 = -\frac{1}{2} \]
The equation \( x^2 = -\frac{1}{2} \) has no real solutions because the square of a real number cannot be negative.
Based on this analysis, the graph of \( f(x) = -2x^2 - 1 \):
- Has a vertex at (0, -1).
- Is a downward-opening parabola.
- Does not intersect the x-axis, as indicated by the negative value for \( x^2 \).
Thus, there are no real solutions because the graph neither touches nor crosses the x-axis.
The correct response is:
no real solutions because the graphs neither touches nor crosses the x-axis.
1 answer