Question 53229
a)  You need to make a perfect square trinomial in such a way that you don't change the value of the function.  Make sure that if you add anything you also take it away.  Right now this is the standard form of a parabola right now.
a){{{y=x^2-4x-5}}}
{{{y=(x^2-4x)-5}}}
To find the number that makes a perfect square take 
{{{((1/2)b)^2}}}
{{{((1/2)(-4))^2}}}
{{{(-4/2)^2}}}
{{{(-2)^2}}}
=4 Add 4 to the inside of the parenthesis and take 4 away from the outside of the parenthesis.
{{{y=(x^2-4x+4)-5-4}}}
{{{y=(x^2-4x+4)-9}}}
Now the parenthesis has a perfect square:
{{{y=(x-2)^2-9}}}
Now this is the vertex form of a parabola!
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b){{{y=a(x-h)^2+k}}} puts the parabolic function in a form that makes the vertex and the axis of symmetry easy to see.
The vertex is (h,k) 
the axis of symmetry is x=h
In our case h=2, so our axis of symmetry is x=2.
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c) Plot the vertex and y intercept (if it's in a reasonable place), if your asked for more points add and take away 1 from the axis of symmetry and plu that into your function to find the y-values.
Vertex=(2,-9)
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y-int: y=(0-2)^2-9
y=4-9=-5
(0,-5)
{{{graph(300,200,-10,10,-12,10,y=(x-2)^2-9)}}}
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d) It's shifted right two units and down 9 units.  Observe:
{{{graph(300,200,-10,10,-12,10,y=(x-2)^2-9,x^2)}}}