Question

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

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

Answer 1

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

Explanation:

On ed :)

Answer 2

`f(x)=ax^2+bx+c`

The graph of this function has minimum if a > 0.

Minimum is in point (h; k) where:


`h=(-b)/(2a);\ k=f(h)`


`minimum(4;-3)\to h=4;\ k=-3`


`f(x)=x^2+4x-11;\ a=1 > 0;\ b = 4\\\\h=(-4)/(2\cdot1)=-2\\eq4`


`f(x)=-2x2+16x-35;\ a=-2 < 0`


`f(x)=x^2-4x+5;\ a=1 > 0;\ b=-4\\\\h=(-(-4))/(2\cdot1)=(4)/(2)=2\\eq4`


`f(x)=2x^2-16x+35;\ a=2 > 0;\ b=-16\\\\h=(-(-16))/(2\cdot2)=(16)/(4)=4\\\\k=f(4)=2\cdot4^2-16\cdot4+35=2\cdot16-64+35=32-64+35=3\\eq-3`

Answer: Any graph of this functions hasn't a minimum located at (4; -3).

If the coordinates of minimum are (4; 3),
then your answer is f(x) = 2x² - 16x + 35