Question

How many real solutions does f(x)=-2x^2-1 have, based on the graph? (1 point) two real solutions because the graph is symmetric about the y-axis one real solution because the graph touches y-axis at y= no real solutions because the graphs neither touches nor crosses the x-axis one real solution because the graph intersects the y-axis once

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Answer
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Answered by GPT-4o mini
To determine how many real solutions the function \( f(x) = -2x^2 - 1 \) has based on its graph, we can analyze its features.

1. **Type of function**: This is a quadratic function of the form \( f(x) = ax^2 + bx + c \), where \( a = -2 \), \( b = 0 \), and \( c = -1 \).

2. **Graph shape**: The coefficient \( a = -2 \) is negative, meaning the parabola opens downwards.

3. **Vertex**: The vertex of this parabola can be found using the formula for the vertex of a quadratic \( x = -\frac{b}{2a} \). Here, \( b = 0 \), so the vertex is at \( x = 0 \). Plugging this back into the function gives:
\[
f(0) = -2(0)^2 - 1 = -1
\]
Therefore, the vertex is at \( (0, -1) \).

4. **Intercepts**: Now, we find the \( y \)-intercept by evaluating \( f(0) \), which we've already calculated to be \( -1 \). So, the point where the graph intersects the \( y \)-axis is \( (0, -1) \).

5. **Finding real solutions**: To find real solutions, we set \( f(x) = 0 \):
\[
-2x^2 - 1 = 0
\]
Rearranging gives:
\[
-2x^2 = 1 \quad \Rightarrow \quad x^2 = -\frac{1}{2}
\]
Since \( x^2 = -\frac{1}{2} \) has no real solutions (as the square of a real number cannot be negative), this implies that the graph does not intersect the \( x \)-axis.

From this analysis, we conclude that the function \( f(x) = -2x^2 - 1 \) has no real solutions since the graph does not touch or cross the \( x \)-axis.

**Answer**: No real solutions because the graph neither touches nor crosses the x-axis.
Answered by GPT-4o mini
Yes, I'm confident in my analysis. Let's break it down again to ensure completeness:

1. **Function Form**: We have the quadratic function \( f(x) = -2x^2 - 1 \).

2. **Direction of the Parabola**: This function opens downwards because the coefficient of \( x^2 \) (which is -2) is negative.

3. **Vertex**: The vertex of the parabola can be found at \( x = -\frac{b}{2a} \). In this case, \( b = 0 \), so:
\[
x = -\frac{0}{2(-2)} = 0
\]
Evaluating \( f(0) \) gives:
\[
f(0) = -2(0)^2 - 1 = -1
\]
Thus, the vertex is at the point \( (0, -1) \).

4. **Y-Intercept**: The graph intersects the y-axis at \( (0, -1) \), confirming the vertex position.

5. **X-Intercepts**: To find the x-intercepts (real solutions), we set \( f(x) = 0 \):
\[
-2x^2 - 1 = 0
\]
Rearranging gives:
\[
-2x^2 = 1 \quad \Rightarrow \quad x^2 = -\frac{1}{2}
\]
Since \( x^2 = -\frac{1}{2} \) does not yield any real solutions (as a square cannot equal a negative number), we conclude that there are indeed **no real solutions** for this function.

6. **Graph Characteristics**: The parabola opens downwards and its vertex is below the x-axis, which means it cannot touch or cross the x-axis anywhere.

Thus, I reaffirm that the correct conclusion is that there are **no real solutions** because the graph neither touches nor crosses the x-axis.

If you have any further questions or need clarification, feel free to ask!

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