Question 408202
{{{f(x) = (1/5)*(x + 2)^2 + 2}}}
I can say {{{z = x + 2}}}, then
{{{f(z) = (1/5)*z^2 + 2}}}
In an equation of the form
{{{f(z) = a*z^2 + b*z + c}}}, the z-coordinate of
the vertex is at {{{-b/(2a)}}}, but {{{b = 0}}}
here, so (0,f(0)) is the vertex, and to find {{{y}}}
{{{f(0) = (1/5)*0^2 + 2}}}
The vertex is at (0,2)
Now I translate this to (x,f(x))
{{{z = x + 2}}}
{{{0 = x + 2}}}
{{{x = -2}}}
{{{f(-2) = (1/5)*(0)^2 + 2}}}
{{{f(-2) = 2}}}
(x,f(x)) = (-2,2)
The line of symmetry is {{{x = -2}}}
The graph is a min at {{{f(x) = 2}}}
Here is the plot:
{{{ graph( 400, 400, -10, 10, -10, 10, (1/5)*(x + 2)^2 + 2) }}}