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Jackson writes the quadratic function f(x)=3(x−10)2−6 . Complete the inequality to show the range of the function.(1 point) The...Asked by 0000
Jackson writes the quadratic function f(x)=3(x−10)^2−6
. Complete the inequality to show the range of the function.(1 point)
The range of the function is given by the inequality y≥
. Complete the inequality to show the range of the function.(1 point)
The range of the function is given by the inequality y≥
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GPT-4o mini
To determine the range of the function \( f(x) = 3(x-10)^2 - 6 \), we first identify its vertex and characteristics.
1. The function \( f(x) = 3(x-10)^2 - 6 \) is in vertex form \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.
2. Here, \( a = 3 \), \( h = 10 \), and \( k = -6 \). Since \( a \) is positive, the parabola opens upwards, meaning the vertex represents the minimum point of the function.
3. The vertex is at the point \( (10, -6) \).
Since the minimum value of \( f(x) \) (the vertex) is \(-6\) and the parabola opens upward, the range of \( f(x) \) includes all values greater than or equal to \(-6\).
Thus, the range of the function is given by the inequality:
\[
y \geq -6
\]
1. The function \( f(x) = 3(x-10)^2 - 6 \) is in vertex form \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.
2. Here, \( a = 3 \), \( h = 10 \), and \( k = -6 \). Since \( a \) is positive, the parabola opens upwards, meaning the vertex represents the minimum point of the function.
3. The vertex is at the point \( (10, -6) \).
Since the minimum value of \( f(x) \) (the vertex) is \(-6\) and the parabola opens upward, the range of \( f(x) \) includes all values greater than or equal to \(-6\).
Thus, the range of the function is given by the inequality:
\[
y \geq -6
\]
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