Question 133003
{{{f(x) = (1/2) (x+4)^2 + 6}}} Start with the given equation



{{{f(x) = (1/2) (x-(-4))^2 + 6}}} Rewrite {{{x+4}}} as {{{x-(-4)}}}



Notice how the equation is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=1/2}}}, {{{h=-4}}} and {{{k=6}}}



Remember, the vertex is (h,k). So in this case the vertex is (-4,6). 


The line of symmetry is in the form {{{x=h}}}. So the equation is {{{x=-4}}}



Since {{{a>0}}} (ie it's positive), the parabola opens up. So this means that a min will occur at the vertex. So in this case the minimum is {{{y=6}}}






If we graph the function, we can visually verify our answer.



{{{drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph( 500, 500, -10, 10, -10, 10, (1/2) (x+4)^2 + 6),
blue(line(-4,-20,-4,20)),
circle(-4,6,0.12),
circle(-4,6,0.14),
circle(-4,6,0.16)
)
 }}} Graph of {{{f(x) = (1/2) (x+4)^2 + 6}}} with the axis of symmetry {{{x=-4}}} (blue line) and the vertex (-4,6)