Question 962351


{{{f(x)= (x-4)^2 -1}}}....compare to vertex form {{{f(x)= (x-h)^2 +k}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the vertex

{{{f(x)= (x-4)^2 -1}}}...as you can see,{{{h=4}}} and {{{-1}}}; so, the vertex is at ({{{4}}},{{{-1}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(4,-1,.12),locate(4,-1,V(4,-1)),
 graph( 600, 600, -10, 10, -10, 10, (x-4)^2 -1)) }}}