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A coordinate plane's x-axis ranges from negative 10 to 2 and its y-axis ranges from negative 4 to 4, both by 1-unit increments. A concave down parabola is drawn passing through the 2nd, 3rd, and 4th quadrants.

Which response is the correct way to write the equation in vertex form for the function?

(1 point)
Responses

y=−14(x+2)2+4
y equals negative Start Fraction 1 over 4 End Fraction left parenthesis x plus 2 right parenthesis squared plus 4

y=−14(x−2)2+4
y equals negative Start Fraction 1 over 4 End Fraction left parenthesis x minus 2 right parenthesis squared plus 4

y=−14(x−4)2+2
y equals negative Start Fraction 1 over 4 End Fraction left parenthesis x minus 4 right parenthesis squared plus 2

y=−14(x+4)2+2

1 answer

To determine the correct equation in vertex form for the parabola described, let's analyze the general form of a vertex equation, which is:

\[ y = a(x - h)^2 + k \]

Here, \((h, k)\) is the vertex of the parabola and \(a\) indicates the orientation (concave up if \(a > 0\) and concave down if \(a < 0\)). The problem states that the parabola is concave down, so \(a\) will be negative.

  1. Vertex Position: Since the parabola passes through the 2nd, 3rd, and 4th quadrants and is concave down, the vertex will likely be located in the 2nd quadrant (where \(x < 0\) and \(y > 0\)).

  2. Identify a Vertex: Given a reasonable guess based on a standard graph, the vertex might likely be around the point \((-2, 4)\) or close, since it is positive in the \(y\) direction and negative in \(x\).

  3. Find 'a': The prompt references \(−14\) and \(-\frac{1}{4}\) for the leading coefficient \(a\). This signifies that the parabola opens downwards and provides some scale for the parabola's stretch.

Given these points:

  • The vertex seems around \((-2, 4)\) based on the description of the parabola being in the 2nd quadrant with the y-value approaching but not exceeding 4.

Considering the vertex, the correct equation would likely be: \[ y = -14(x + 2)^2 + 4 \]

As a result, the most fitting response would be: y = -14(x + 2)^2 + 4.

This matches the vertex at \((-2, 4)\) while maintaining a downwards-opening parabola.