Question

1. identify the quadratic equation for the graph by determining the vertex and creating a table of values.options: y = -x² + 4; y = -x² - 4; v = -(x + 4)²

Explanation:

Step1: Identify the vertex

The vertex of the parabola is at \((0, 4)\) (since it's on the y - axis, \(x = 0\) and the y - coordinate is 4).

Step2: Analyze the form of quadratic equations

The vertex form of a quadratic equation is \(y=a(x - h)^2+k\), where \((h,k)\) is the vertex. For our vertex \((0,4)\), \(h = 0\) and \(k = 4\), so the equation becomes \(y=a(x-0)^2 + 4=a x^2+4\).

Step3: Determine the sign of \(a\)

The parabola opens downward, so \(a<0\).

Step4: Check the options

• For option \(y=-x^2 + 4\): It is in the form \(y = ax^2+4\) with \(a=- 1<0\) (opens downward) and the vertex at \((0,4)\), which matches our analysis.
• For option \(y=-x^2-4\): The vertex would be at \((0,-4)\), which does not match the vertex \((0,4)\) we identified from the graph.
• For option \(y=-(x + 4)^2\): The vertex form is \(y=a(x - h)^2+k\) with \(h=-4\) and \(k = 0\), so the vertex is at \((-4,0)\), which does not match the vertex \((0,4)\) from the graph.

Answer:

A. \( y = -x^2 + 4 \)