Question

The graph of which function has a minimum located at (4, –3)?

f(x) = one-halfx2 + 4x – 11
f(x) = –2x2 + 16x – 35
f(x) = one-halfx2 – 4x + 5
f(x) = 2x2 – 16x + 35

Answer 1

Answer:

f(x) = 1/2x^2 – 4x + 5  (the third option)

Step-by-step explanation:

I looked it up on a graphing calculator.

Answer 2

The quadratic function that has a minimum located at (4, -3) is:

$f(x) = (1/2)*x^2 - 4x + 5$

Which function has a minimum located at (4, -3)?

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:

$f(x) = (1/2)*x^2 + 4x -11$

$f(x) = (1/2)*x^2 - 4x + 5$

$f(x) = 2x^2 - 16x + 35$

The vertex of the first one is at:

$x = -4/(2*1/2) = -4$

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:

$x = -(-4)/(2*1/2) = 4$

Then the y-value of the vertex is:

$f(4) = (1/2)*4^2 - 4*4 + 5 = -3$

Then the vertex is at (4, -3), so this is the correct option.

If you want to learn more about quadratic functions:

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