Question 59767
1. {{{ y = x^2 + 5 }}}
2. {{{ y = 2x + 4 }}}
Since both are already in "y=" .... set them equal to each other
{{{ x^2 + 5 = 2x + 4 }}}
Move all parts to the left ... setting the equation equal to zero
{{{ x^2 +5 -2x -4 = 0 }}}
combine like terms and put in descending order
{{{ x^2 -2x +1 = 0 }}}
Factor
{{{ (x-1)(x-1) = 0 }}}
Since both parts are identical ... we only need to work it once
{{{ x-1 = 0 }}}
{{{ x = 1 }}}


Pulg x = 1 into the equations
{{{ y = x^2 + 5 }}} and {{{ y = 2x + 4 }}}
{{{ y = 1^2 + 5 }}} and {{{ y = 2(1) + 4 }}}
{{{ y = 1 + 5 }}} and {{{ y = 2 + 4 }}}
{{{ y = 6 }}} and {{{ y = 6 }}}


Your one solution is 
(1,6)

The graphs are a parabola and a line.  Since there is only one solution, the point at which they meet is also the "minimum" or vertex of the parabola.