To graph the equation \( y = -5x^2 + 1 \), we can start by identifying key features of the quadratic function.
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Shape: This is a downward-opening parabola because the coefficient of \( x^2 \) is negative (-5).
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Vertex: The vertex of the parabola can be found using the vertex form or by completing the square. Here, since it’s already in the form \( y = ax^2 + bx + c \), the vertex occurs at the maximum point. In standard form \( y = ax^2 + c \) (with \( b = 0 \)), the vertex is at the point \( (0, c) \). Thus, the vertex for this equation is at \( (0, 1) \).
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Y-intercept: The y-intercept occurs where \( x = 0 \), which gives us \( y = -5(0)^2 + 1 = 1 \). So, the y-intercept is \( (0, 1) \).
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X-intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = -5x^2 + 1 \ 5x^2 = 1 \ x^2 = \frac{1}{5} \ x = \pm \frac{1}{\sqrt{5}} \approx \pm 0.447 \]
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Direction: The parabola opens downward due to the negative coefficient of \( x^2 \).
To summarize about the graph:
- The vertex is at \( (0, 1) \).
- The parabola opens downwards.
- The y-intercept is at \( (0, 1) \).
- The x-intercepts are approximately \( (0.447, 0) \) and \( (-0.447, 0) \).
If you have specific statements to evaluate, please provide them, and I'll help determine which is correct based on the analysis above!