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Question

A downward open parabola, made of a solid line is plotted on a coordinate plane. The x-axis ranges from negative 5 to 5 and the y-axis ranges from 0 to negative 10, both in increments of 5.

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

(1 point)
Responses

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

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

two real solutions because the graph is symmetric about the y-axis
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
no real solutions because the graphs neither touches nor crosses the x -axis
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Answer 1

To determine the number of real solutions for the equation \( f(x) = -2x^2 - 1 \), we should analyze what it means for a parabola to have real solutions.

Given that this is a downward-opening parabola, the vertex of the parabola represents the maximum point, and the equation suggests its y-intercept is at \( -1 \) when \( x = 0 \).

For the parabola to have real solutions (x-intercepts), it must touch or cross the x-axis. In this case, since the vertex is the highest point of the graph and is located at \( y = -1 \), which is below the x-axis, and as the parabola opens downward, it will not cross the x-axis at any point.

Thus, the correct conclusion is:

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