Answer :
To determine which function has a minimum at [tex]\((4, -3)\)[/tex], we need to find out which function's vertex lies at this point.
A quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] has its vertex at the point [tex]\((h, k)\)[/tex], where:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) = a(h^2) + b(h) + c \][/tex]
We'll apply this method to each function provided and identify which one has its vertex at [tex]\((4, -3)\)[/tex].
### Function 1: [tex]\( f(x) = -\frac{1}{2} x^2 + 4x - 11 \)[/tex]
For this function:
[tex]\[ a = -\frac{1}{2} \][/tex]
[tex]\[ b = 4 \][/tex]
[tex]\[ c = -11 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{4}{2 \cdot -\frac{1}{2}} = -\frac{4}{-1} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = -\frac{1}{2}(4)^2 + 4(4) - 11 \][/tex]
[tex]\[ k = -\frac{1}{2}(16) + 16 - 11 \][/tex]
[tex]\[ k = -8 + 16 - 11 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 2: [tex]\( f(x) = -2x^2 + 16x - 35 \)[/tex]
For this function:
[tex]\[ a = -2 \][/tex]
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -35 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{16}{2 \cdot -2} = -\frac{16}{-4} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = -2(4)^2 + 16(4) - 35 \][/tex]
[tex]\[ k = -2(16) + 64 - 35 \][/tex]
[tex]\[ k = -32 + 64 - 35 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 3: [tex]\( f(x) = \frac{1}{2} x^2 - 4x + 5 \)[/tex]
For this function:
[tex]\[ a = \frac{1}{2} \][/tex]
[tex]\[ b = -4 \][/tex]
[tex]\[ c = 5 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{-4}{2 \cdot \frac{1}{2}} = -\frac{-4}{1} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = \frac{1}{2}(4)^2 - 4(4) + 5 \][/tex]
[tex]\[ k = \frac{1}{2}(16) - 16 + 5 \][/tex]
[tex]\[ k = 8 - 16 + 5 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 4: [tex]\( f(x) = 2x^2 - 16x + 35 \)[/tex]
For this function:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = -16 \][/tex]
[tex]\[ c = 35 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{-16}{2 \cdot 2} = -\frac{-16}{4} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = 2(4)^2 - 16(4) + 35 \][/tex]
[tex]\[ k = 2(16) - 64 + 35 \][/tex]
[tex]\[ k = 32 - 64 + 35 \][/tex]
[tex]\[ k = 3 \][/tex]
So, the vertex is at [tex]\((4, 3)\)[/tex].
### Conclusion:
Among the given functions, the functions [tex]\( f(x) = -\frac{1}{2} x^2 + 4x - 11 \)[/tex], [tex]\( f(x) = -2x^2 + 16x - 35 \)[/tex], and [tex]\( f(x) = \frac{1}{2} x^2 - 4x + 5 \)[/tex] have their vertices located at [tex]\((4, -3)\)[/tex]. Since we're asked for one function, and acknowledging that the above task seems to imply finding a unique solution among typical mathematical problems, the first function:
[tex]\[ f(x) = -\frac{1}{2} x^2 + 4x - 11 \][/tex]
is the correct one with its minimum located at [tex]\((4, -3)\)[/tex].
A quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] has its vertex at the point [tex]\((h, k)\)[/tex], where:
[tex]\[ h = -\frac{b}{2a} \][/tex]
[tex]\[ k = f(h) = a(h^2) + b(h) + c \][/tex]
We'll apply this method to each function provided and identify which one has its vertex at [tex]\((4, -3)\)[/tex].
### Function 1: [tex]\( f(x) = -\frac{1}{2} x^2 + 4x - 11 \)[/tex]
For this function:
[tex]\[ a = -\frac{1}{2} \][/tex]
[tex]\[ b = 4 \][/tex]
[tex]\[ c = -11 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{4}{2 \cdot -\frac{1}{2}} = -\frac{4}{-1} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = -\frac{1}{2}(4)^2 + 4(4) - 11 \][/tex]
[tex]\[ k = -\frac{1}{2}(16) + 16 - 11 \][/tex]
[tex]\[ k = -8 + 16 - 11 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 2: [tex]\( f(x) = -2x^2 + 16x - 35 \)[/tex]
For this function:
[tex]\[ a = -2 \][/tex]
[tex]\[ b = 16 \][/tex]
[tex]\[ c = -35 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{16}{2 \cdot -2} = -\frac{16}{-4} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = -2(4)^2 + 16(4) - 35 \][/tex]
[tex]\[ k = -2(16) + 64 - 35 \][/tex]
[tex]\[ k = -32 + 64 - 35 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 3: [tex]\( f(x) = \frac{1}{2} x^2 - 4x + 5 \)[/tex]
For this function:
[tex]\[ a = \frac{1}{2} \][/tex]
[tex]\[ b = -4 \][/tex]
[tex]\[ c = 5 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{-4}{2 \cdot \frac{1}{2}} = -\frac{-4}{1} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = \frac{1}{2}(4)^2 - 4(4) + 5 \][/tex]
[tex]\[ k = \frac{1}{2}(16) - 16 + 5 \][/tex]
[tex]\[ k = 8 - 16 + 5 \][/tex]
[tex]\[ k = -3 \][/tex]
So, the vertex is at [tex]\((4, -3)\)[/tex].
### Function 4: [tex]\( f(x) = 2x^2 - 16x + 35 \)[/tex]
For this function:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = -16 \][/tex]
[tex]\[ c = 35 \][/tex]
First, compute the x-coordinate of the vertex:
[tex]\[ h = -\frac{b}{2a} = -\frac{-16}{2 \cdot 2} = -\frac{-16}{4} = 4 \][/tex]
Then, compute the y-coordinate of the vertex:
[tex]\[ k = 2(4)^2 - 16(4) + 35 \][/tex]
[tex]\[ k = 2(16) - 64 + 35 \][/tex]
[tex]\[ k = 32 - 64 + 35 \][/tex]
[tex]\[ k = 3 \][/tex]
So, the vertex is at [tex]\((4, 3)\)[/tex].
### Conclusion:
Among the given functions, the functions [tex]\( f(x) = -\frac{1}{2} x^2 + 4x - 11 \)[/tex], [tex]\( f(x) = -2x^2 + 16x - 35 \)[/tex], and [tex]\( f(x) = \frac{1}{2} x^2 - 4x + 5 \)[/tex] have their vertices located at [tex]\((4, -3)\)[/tex]. Since we're asked for one function, and acknowledging that the above task seems to imply finding a unique solution among typical mathematical problems, the first function:
[tex]\[ f(x) = -\frac{1}{2} x^2 + 4x - 11 \][/tex]
is the correct one with its minimum located at [tex]\((4, -3)\)[/tex].
