How many real solutions does f(x)=−2x2−1 have, based on the graph?(1 point)Responsesno real solutions because the graphs neither touches nor crosses the x-axisno real solutions because the graphs neither touches nor crosses the x -axistwo real solutions because the graph is symmetric about the y-axistwo real solutions because the graph is symmetric about the y -axisone real solution because the graph intersects the y-axis once one real solution because the graph intersects the y -axis once one real solution because the graph touches y-axis at y=−1one real solution because the graph touches y -axis at y equals negative 1

Answer 1

To determine the number of real solutions for the function \( f(x) = -2x^2 - 1 \), we need to analyze its graph.

This function is a downward-opening parabola because the coefficient of \( x^2 \) is negative (\( -2 \)). The constant term is \( -1 \), which means that the vertex of the parabola is located at \( (0, -1) \).

To find the x-intercepts (i.e., where the graph intersects the x-axis), we set \( f(x) = 0 \):

\[ -2x^2 - 1 = 0 \]

Rearranging this gives:

\[ -2x^2 = 1 \]

\[ x^2 = -\frac{1}{2} \]

Since \( x^2 \) cannot be negative for any real \( x \), there are no real solutions to \( f(x) = 0 \).

This means that the graph of \( f(x) \) does not touch or cross the x-axis, indicating that there are no real solutions to this equation.

Therefore, the answer is:

No real solutions because the graph neither touches nor crosses the x-axis.