Question 407300
Put y for f(x) and get 
y = (x+3)^2 +1
y = x^2 + 6*x +9 +1
y = x^2 + 6*x + 10
we plot a group of quadratic equation (x,y) points for a group of x say {0, 1,-1,2,-2,3,-3}.
if x=0 then y = x^2 + 6*0 +10 so y=10 and (0,10) is a solution point for the quadratic equation y = x^2 + 6*x + 10.
For x=-3 y=3*2+6*3+10=17 and (-3,1) is a point and similarly for 
x=-1,x=2,x=-2,x=3,x=-3
{{{ graph( 500, 500, -6, 5, -10, 10, 0, x^2+6*x+10) }}}