Question 518509
find the vertex, the line of symetry, and the maximum or minimum value of f(x). graph
<pre>
Rules:

For f(x) = a(x-h)² + k

vertex = (h,k)

line of symetry = the line whose equation is x = h, a vertical line
through h on the x-axis: 

max/min value = k 

Is it max or min = max?  Rule: if a > 0 it is a min, if a < 0, it is
a max.  

To draw the graph:  

1. Draw the line of symmetry.
2. Plot the vertex (h,k), 
3. Plot the two points (h+1,k+a), (h-1,k+a)
4. Draw a u-shaped curve through them, symmetrical about the line
of symmetry.

-------------------------------
Use the rule on your equation:

f(x) = {{{1/3}}}(x+5)² + 6

Compare to:

f(x) = a(x-h)² + k

a = {{{1/3}}}, h = -5, k = 6

vertex = (h,k) = (-5,6)

line of symetry = the line whose equation is x = h, or x = -5 

max/min value = k = 6 

Is it max or min = max?  Rule: if a > 0 it is a min, if a < 0, it is
a max.   Since a = {{{1/3}}} which is > 0, it is a min.
  
To draw the graph:  

1. Draw the line of symmetry, a vertical line through -5 on the x-axis,
whose equation is x = -5

{{{drawing(400,400,-12,2,-2,12, graph(400,400,-12,2,-2,12),
green(line(-5,-3,-5,13)) )}}}



2. Plot the vertex (h,k) = (-5,6)

{{{drawing(400,400,-12,2,-2,12, graph(400,400,-12,2,-2,12),
green(line(-5,-3,-5,13)), circle(-5,6,.1), locate(-5,6,"(-5,6)") )}}}




3. Plot the two points (h+1,k+a) = (-5+1,6+{{{1/3}}}) = (-4,{{{6&1/3}}})
and (h-1,k+a) = (-5-1,6+{{{1/3}}}) = (-6,{{{6&1/3}}})

{{{drawing(400,400,-12,2,-2,12, graph(400,400,-12,2,-2,12),
green(line(-5,-3,-5,13)), circle(-5,6,.1), locate(-5,6,"(-5,6)"),

locate(-9+.2,6+.2,"(-6,6"), locate(-5,6,"(-5,6)"),
locate(-7.5+.2,6.4+.2,2/3), locate(-7+.2,6+.2,")"), circle(-6,6+1/3,.1),

locate(-9+.2+5,6+.4,"(-4,6"), 
locate(-7.5+5+.4,6.4+.4,2/3), locate(-7+.2+5,6+.4,")"), circle(-4,6+1/3,.1)
)}}}

4. Draw a u-shaped curve through them, symmetrical about the line
of symmetry.

{{{drawing(400,400,-12,2,-2,12, graph(400,400,-12,2,-2,12, (1/3)(x+5)^2+6),
green(line(-5,-3,-5,13)), circle(-5,6,.1), locate(-5,6,"(-5,6)"),

locate(-9+.2,6+.2,"(-6,6"), locate(-5,6,"(-5,6)"),
locate(-7.5+.2,6.4+.2,2/3), locate(-7+.2,6+.2,")"), circle(-6,6+1/3,.1),

locate(-9+.2+5,6+.4,"(-4,6"), 
locate(-7.5+5+.4,6.4+.4,2/3), locate(-7+.2+5,6+.4,")"), circle(-4,6+1/3,.1)
)}}}

Edwin</pre>