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

The parabola with the minimum at (4, -3) is the third option:

$y = 0.5*x^2 - 4x + 5$

Which of the given parabolas has a minimum at (4, -3)?

Remember that for a parabola of the form:

$y = a*x^2 + b*x + c$

If a > 0, the minimum is at the x-value of the vertex:

$x = \frac{-b}{2a}$

If a < 0, there is no minimum (so we discard option 2).

Taking the third option:

$y = 0.5*x^2 - 4x + 5$

The x-value of the vertex is:

$x = \frac{-(-4)}{2*0.5} = 4$

As expected.

To get the y-value of the vertex (and minimum) we just evaluate the parabola equation in x = 4.

$y = 0.5*(4)^2 - 4*4 + 5 = -3$

So the minimum is at (4, -3), as expected, so that is the correct option.

If you want to learn more about parabolas:

https://brainly.com/question/4061870

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