[tex]f(x)=ax^2+bx+c[/tex]
The graph of this function has minimum if a > 0.
Minimum is in point (h; k) where:
[tex]h=\dfrac{-b}{2a};\ k=f(h)[/tex]
[tex]minimum(4;-3)\to h=4;\ k=-3[/tex]
[tex]f(x)=x^2+4x-11;\ a=1 > 0;\ b = 4\\\\h=\dfrac{-4}{2\cdot1}=-2\neq4[/tex]
[tex]f(x)=-2x2+16x-35;\ a=-2 < 0[/tex]
[tex]f(x)=x^2-4x+5;\ a=1 > 0;\ b=-4\\\\h=\dfrac{-(-4)}{2\cdot1}=\dfrac{4}{2}=2\neq4[/tex]
[tex]f(x)=2x^2-16x+35;\ a=2 > 0;\ b=-16\\\\h=\dfrac{-(-16)}{2\cdot2}=\dfrac{16}{4}=4\\\\k=f(4)=2\cdot4^2-16\cdot4+35=2\cdot16-64+35=32-64+35=3\neq-3[/tex]
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
