Question 252576
I am trying to find out two symmetrical points from the equation y= -1/3(x-5)^2 -8 the vertex is 5,-8 and the symmetry axis is x=5 the parabola is concave down as -1/3 is a negative. I am sruggling to find the two points so i can do a graph.
<pre><font size = 4 color = "indigo"><b>
(-1,-20) and (11,-20 are symmetrical points
 (2,-11) and (8,-11) are symmetrical points

{{{drawing(400,685.7142857,-2,12,-22,2, 

line(-1+.2,-20,-1-.2,-20), line(-1,-20+.2,-1,-20-.2), line(-1+.2,-20+.2,-1-.2,-20-.2), line(-1+.2,-20-.2,-1-.2,-20+.2), green(line(5,-23,5,3)),
line(11+.2,-20,11-.2,-20), line(11,-20+.2,11,-20-.2), line(11+.2,-20+.2,11-.2,-20-.2), line(11+.2,-20-.2,11-.2,-20+.2),  
line(2+.2,-11,2-.2,-11),line(2,-11+.2,2,-11-.2),line(2+.2,-11-.2,2-.2,-11+.2),line(2-.2,-11+.2,2+.2,-11-.2),
line(8+.2,-11,8-.2,-11), line(8,-11+.2,8,-11-.2),line(8+.2,-11+.2,8-.2,-11-.2),line(8+.2,-11-.2,8-.2,-11+.2), locate(2,-11,"(2,-11)"),

locate(-1,-20,"(-1,20)"),  locate(8.5,-20,"(11,-20)"), locate(8.5,-11,"(8,-11)"),

graph(400,685.7142857,-2,12,-22,2, -(1/3)(x-5)^2-8) )}}}
 
Edwin</pre>