Question

select the graph of the equation below.
$y = -\frac{1}{4}x^2 + 1$
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.

Explanation:

Step1: Analyze the vertex form

The equation \( y = -\frac{1}{4}x^2 + 1 \) is a quadratic in vertex form \( y = ax^2 + k \), where the vertex is at \( (0, k) \). Here, \( k = 1 \), so the vertex is \( (0, 1) \).

Step2: Check the y - intercept

When \( x = 0 \), \( y = -\frac{1}{4}(0)^2 + 1 = 1 \), so the graph passes through \( (0, 1) \).

Step3: Analyze the direction and width

The coefficient \( a = -\frac{1}{4} \) is negative, so the parabola opens downward. The absolute value of \( a \) is less than 1, so it is wider than the standard parabola \( y = x^2 \).

Step4: Compare with the graphs

• Graph A: The vertex is at \( (0, 1) \), opens downward, and has the correct width.
• Graph B: The vertex is at \( (0, 0) \) (or close to it), which does not match the y - intercept of 1.

Answer:

A. The graph with vertex at (0,1), opening downward, passing through (0,1) and symmetric about the y - axis.