Answer
Vertex: $(-5,0)$
x-intercept: $(-5,0)$
y-intercept: $(0, 25)$
Work Step by Step
$f(x)=x^2+10x+25$
$a=1$, $b=10$, $c=25$
Vertex is at $x=-b/2a$
$x=-10/2*1$
$x=-10/2$
$x=-5$
$f(x)=x^2+10x+25$
$f(-5)=(-5)^2+10*(-5)+25$
$f(-5)=25-50+25$
$f(-5)=0$
$(-5,0)$ is the vertex.
$f(x)=x^2+10x+25$
$y=x^2+10x+25$
$y=0$
$y=x^2+10x+25$
$0=x^2+10x+25$
$0=(x+5)^2$
$\sqrt 0 = \sqrt {(x+5)^2}$
$0 = (x+5)$
$0-5=x+5-5$
$-5=x$
$x=0$
$y=x^2+10x+25$
$y=0^2+10*0+25$
$y=0+0+25$
$y=25$
$(0,25)$
