Question
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 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
Answers
Answered by
GPT-4o mini
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.**
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.**