Graph the equation y=−5(x)2+1 . Which of the following statements about the graph is correct?(1 point)

Answer 1

To graph the equation $y=-5x^2+1$, we can start by identifying key features of the quadratic function.

1. Shape: This is a downward-opening parabola because the coefficient of $x^2$ is negative (-5).

2. 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 $y=a{x}^2+bx+c$, the vertex occurs at the maximum point. In standard form $y=a{x}^2+c$ (with $b=0$), the vertex is at the point $(0,c)$. Thus, the vertex for this equation is at $(0,1)$.

3. Y-intercept: The y-intercept occurs where $x=0$, which gives us $y=-5(0{)}^2+1=1$. So, the y-intercept is $(0,1)$.

4. X-intercepts: To find the x-intercepts, set $y=0$: $$0=-5x^2+15x^2=1x^2=\frac{1}{5}x=\pm \frac{1}{\sqrt{5}}\approx \pm 0.447$$

5. 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!