Question 137704
y = (x-3)^2 - 5
:
Solve by making y = 0
(x-3)^2 - 5 = 0
:
(x-3)^2 = +5; added 5 to both sides
:
x - 3 = +/-{{{sqrt(5)}}};find the square root of both sides
:
x = 3 + {{{sqrt(5)}}}; added 3 to both sides
and 
x = 3 - {{{sqrt(5)}}}
:
When you graph, you will see these are the values for the x intercept
:
Graph using the original equation; Find the value of y for x = -1 to x = +6
y = (-1-3)^2 - 5
y = -4^2 - 5
y = 16 - 5
y = +11
the first entry on the table will be:
 x | y
-------
-1 + +11
:
For x = 0
y = (0-3)^2 - 5
y = +9 - 5
y = +4
the table is now:
 x | y
------
-1 |+11
 0 |+4
:
Do this for x = +1, 2, 3, 4, 5, 6
Your table should look like this
 x | y
------
-1 |+11
 0 |+4
+1 |-1
+2 |-4
+3 |-5
+4 |-4
+5 |-1
+6 |+4
:
The graph should look like this
{{{ graph( 300, 200, -3, 12, -8, 10, (x-3)^2-5) }}}
Note they cross the x axis at about +.76 and +5.24 which are decimal values for 
3-Sqrt(5) and 3+sqrt(5)
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