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

Answer 1

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.

Answer 2

Are you sure

Answer 3

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!