Answer
Vertex: $(-5, -16/5)$
Opens upward
x-intercepts: -9, -1
y-intercept: 9/5
Work Step by Step
$f(x)=1/5x^2+2x+9/5$
$y=1/5x^2+2x+9/5$
$y=(1/5x^2+2x)+9/5$
$y=1/5(x^2+10x)+9/5$
$y=1/5(x^2+10x+(10/2)^2)+9/5-(1/5)(10/2)^2$
$y=1/5(x^2+10x+(5)^2)+9/5-(1/5)(5)^2$
$y=1/5(x^2+10x+25)+9/5-(1/5)(25)$
$y=1/5(x+5)^2+9/5-5$
$y=1/5(x+5)^2-16/5$
Vertex: $(-5, -16/5)$
Opens upward
$x=0$
$f(x)=1/5x^2+2x+9/5$
$f(0)=1/5*0^2+2*0+9/5$
$f(0)=1/5*0+0+9/5$
$f(0)=0+9/5$
$f(0)=9/5$
$y=0$
$y=1/5(x+5)^2-16/5$
$0=1/5(x+5)^2-16/5$
$0+16/5=1/5(x+5)^2-16/5+16/5$
$16/5=1/5(x+5)^2$
$16/5*5=1/5(x+5)^2*5$
$16=(x+5)^2$
$\sqrt {16} = \sqrt {(x+5)^2}$
$±4 = x+5$
$4=x+5$
$4-5=x+5-5$
$-1=x$
$-4=x+5$
$-4-5=x+5-5$
$-9=x$
x-intercepts: -9, -1
y-intercept: 9/5
