Question 1103410
Here's the plot:
{{{ graph( 400,400, -5, 5,-5, 15, 5*( x+1 )^2 + 4 ) }}}
{{{ f(x) = 5*( x+1 )^2 + 4 }}}
{{{ f(x) = 5*( x^2 + 2x + 1 ) + 4 }}}
{{{ f(x) = 5x^2 + 10x + 5 + 4 }}}
{{{ f(x) = 5x^2 + 10x + 9  }}}
the x-value of the vertex is:
{{{ x[v] = -b/(2a) }}}
{{{ a = 5 }}}
{{{ b = 10 }}}
{{{ x[v] = (-10)/( 2*5 ) }}}
{{{ x[v] = -1 }}}
Plug this back into equation
{{{ f(x) = 5*( -1 + 1 ) + 4 }}}
{{{ f(x) = 4 }}}
So, ther vertex is at ( -1, 4 )
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There are no asymptotes with a parabola