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what is the \\(y\\)-value of the vertex of the function \\(f(x)=-(x+8)(…

Question

what is the \\(y\\)-value of the vertex of the function \\(f(x)=-(x+8)(x-14)\\)?

3
6
112
121

Explanation:

Response

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"Intercept Form"
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<reasoning>

Identify the x-intercepts of the function

The given quadratic function is in intercept (or factored) form:
\[f(x) = -(x + 8)(x - 14)\]
The x-intercepts occur where \(f(x) = 0\). Setting each factor to zero gives:
\[x + 8 = 0 \implies x = -8\]
\[x - 14 = 0 \implies x = 14\]

Find the x-coordinate of the vertex

The x-coordinate of the vertex, \(h\), lies exactly halfway between the two x-intercepts due to the symmetry of the parabola:
\[h = \frac{-8 + 14}{2}\]
\[h = \frac{6}{2} = 3\]

Calculate the y-value of the vertex

To find the y-value of the vertex, \(k\), substitute the x-coordinate \(x = 3\) back into the original function:
\[k = f(3) = -(3 + 8)(3 - 14)\]
\[k = -(11)(-11)\]
\[k = -(-121) = 121\]
The y-value of the vertex is \(121\).
</reasoning>

<answer>
<mcq-option>3</mcq-option>
<mcq-option>6</mcq-option>
<mcq-option>112</mcq-option>
<mcq-correct>121</mcq-correct>
</answer>

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Answer:

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<reasoning>

Identify the x-intercepts of the function

The given quadratic function is in intercept (or factored) form:
\[f(x) = -(x + 8)(x - 14)\]
The x-intercepts occur where \(f(x) = 0\). Setting each factor to zero gives:
\[x + 8 = 0 \implies x = -8\]
\[x - 14 = 0 \implies x = 14\]

Find the x-coordinate of the vertex

The x-coordinate of the vertex, \(h\), lies exactly halfway between the two x-intercepts due to the symmetry of the parabola:
\[h = \frac{-8 + 14}{2}\]
\[h = \frac{6}{2} = 3\]

Calculate the y-value of the vertex

To find the y-value of the vertex, \(k\), substitute the x-coordinate \(x = 3\) back into the original function:
\[k = f(3) = -(3 + 8)(3 - 14)\]
\[k = -(11)(-11)\]
\[k = -(-121) = 121\]
The y-value of the vertex is \(121\).
</reasoning>

<answer>
<mcq-option>3</mcq-option>
<mcq-option>6</mcq-option>
<mcq-option>112</mcq-option>
<mcq-correct>121</mcq-correct>
</answer>

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