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To determine the range of the function [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex], let's analyze its different parts step by step.
1. Identify the type of function: The given function is a quadratic function in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], where [tex]\( a \)[/tex], [tex]\( h \)[/tex], and [tex]\( k \)[/tex] are constants. For [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex]:
- The coefficient [tex]\( a \)[/tex] is [tex]\(-1\)[/tex].
- The term [tex]\((x + 3)^2\)[/tex] indicates that [tex]\( h = -3 \)[/tex].
- The constant [tex]\( +7 \)[/tex] indicates that [tex]\( k = 7 \)[/tex].
2. Determine the vertex: For any quadratic function in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], the vertex is the point [tex]\((h, k)\)[/tex]. For the function [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex], the vertex is at [tex]\( (-3, 7) \)[/tex].
3. Open direction of the parabola:
- Since the coefficient of the [tex]\((x+3)^2\)[/tex] term is negative ([tex]\(a = -1\)[/tex]), the parabola opens downwards.
4. Finding the range:
- When a parabola opens downwards, the vertex represents the highest point on the graph.
- Therefore, the maximum value of the function [tex]\( f(x) \)[/tex] is the y-coordinate of the vertex, which is [tex]\( 7 \)[/tex].
Given that the parabola opens downwards from the vertex at [tex]\( (-3, 7) \)[/tex], the function can take any value less than or equal to [tex]\( 7 \)[/tex]. That is:
[tex]\[ f(x) \leq 7 \][/tex]
Hence, the range of the function [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex] is:
[tex]\[ \text{All real numbers less than or equal to 7} \][/tex]
1. Identify the type of function: The given function is a quadratic function in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], where [tex]\( a \)[/tex], [tex]\( h \)[/tex], and [tex]\( k \)[/tex] are constants. For [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex]:
- The coefficient [tex]\( a \)[/tex] is [tex]\(-1\)[/tex].
- The term [tex]\((x + 3)^2\)[/tex] indicates that [tex]\( h = -3 \)[/tex].
- The constant [tex]\( +7 \)[/tex] indicates that [tex]\( k = 7 \)[/tex].
2. Determine the vertex: For any quadratic function in the form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], the vertex is the point [tex]\((h, k)\)[/tex]. For the function [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex], the vertex is at [tex]\( (-3, 7) \)[/tex].
3. Open direction of the parabola:
- Since the coefficient of the [tex]\((x+3)^2\)[/tex] term is negative ([tex]\(a = -1\)[/tex]), the parabola opens downwards.
4. Finding the range:
- When a parabola opens downwards, the vertex represents the highest point on the graph.
- Therefore, the maximum value of the function [tex]\( f(x) \)[/tex] is the y-coordinate of the vertex, which is [tex]\( 7 \)[/tex].
Given that the parabola opens downwards from the vertex at [tex]\( (-3, 7) \)[/tex], the function can take any value less than or equal to [tex]\( 7 \)[/tex]. That is:
[tex]\[ f(x) \leq 7 \][/tex]
Hence, the range of the function [tex]\( f(x) = -(x+3)^2 + 7 \)[/tex] is:
[tex]\[ \text{All real numbers less than or equal to 7} \][/tex]
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