Question 1128688


 {{{f(x)=x^2+25}}}, => parabola, opening up, has minimum

to find the Vertex, find minimum; set {{{x=0}}}
{{{f(x)=0^2 + 25 }}}
{{{f(x)=25 }}}
 
minimum or vertex is at ({{{0}}},{{{25}}})


and to find the {{{x}}}-intercept(s), set {{{f(x)=0}}}

 {{{0=x^2+25}}}

 {{{-25=x^2}}}

{{{x=sqrt(-25)}}}

{{{x=-5i}}} or {{{x=5i}}}=> there is no {{{x}}}-intercept(s)



 {{{ graph(600, 600, -30, 30, -30, 30, x^2+25) }}}