Answer:
f(x) = 1/2x^2 – 4x + 5 (the third option)
Step-by-step explanation:
I looked it up on a graphing calculator.
The quadratic function that has a minimum located at (4, -3) is:
[tex]f(x) = (1/2)*x^2 - 4x + 5[/tex]
All the options are quadratic functions, so we need to find the one with a leading coefficient positive, which also has a vertex at (4, -3).
The 3 ones with positive leading coefficients are:
[tex]f(x) = (1/2)*x^2 + 4x -11[/tex]
[tex]f(x) = (1/2)*x^2 - 4x + 5[/tex]
[tex]f(x) = 2x^2 - 16x + 35[/tex]
The vertex of the first one is at:
[tex]x = -4/(2*1/2) = -4[/tex]
The x-value of the vertex must be x = 4, so we can discard this.
For the second quadratic function the x-value of the vertex is:
[tex]x = -(-4)/(2*1/2) = 4[/tex]
Then the y-value of the vertex is:
[tex]f(4) = (1/2)*4^2 - 4*4 + 5 = -3[/tex]
Then the vertex is at (4, -3), so this is the correct option.
If you want to learn more about quadratic functions:
https://brainly.com/question/1214333
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