How many real solutions does f(x)=−2x2−1 have, based on the graph?

(1 point)
Responses

one real solution because the graph intersects the y -axis once

two real solutions because the graph is symmetric about the y -axis

no real solutions because the graphs neither touches nor crosses the x-axis

one real solution because the graph touches y-axis at y=−1

1 answer

To determine how many real solutions the equation \( f(x) = -2x^2 - 1 \) has, we need to analyze the function.

  1. The function is a quadratic function in standard form \( f(x) = ax^2 + bx + c \), where in this case \( a = -2 \), \( b = 0 \), and \( c = -1 \).

  2. The leading coefficient \( a = -2 \) is negative, indicating that the parabola opens downwards.

  3. 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.

  4. 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 \( (0, -1) \).

  5. 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.