Question 921157
<pre>
Maybe the other tutor didn't break it down enough for you 
to understand.  Some people require a little more breakdown 
and explanation than others.  Here is the complete breakdown 
all the way to the graph.  Study it carefully and use it as
a model for your other problems and you'll know how to do them
all.

{{{y = 4x^2+24x-13}}}

Factor 4 out of the first two terms. (Don't factor out 4x, just 4)

{{{y = 4(x^2+6x[""])-13}}}

To the side:
Multiply the coefficient of x, which is 6, by {{{1/2}}}.  {{{6*expr(1/2)=3}}}
Square 3.   {{{3^2=9}}}
Add +9-9 to the 6x inside the parentheses. (It doesn't affect the value
because adding +9-9 is the same as adding 0.

{{{y = 4(x^2+6x+9-9[""])-13}}}

To the side:
Factor the first three terms inside the parentheses: x²+6x+9 = (x+3)(x+3)
Write (x+3)(x+3) as (x+3)². 
Replace x²+6x+9 by (x+3)²

{{{y = 4((x+3)^2-9[""])-13}}}

Remve the BIG parentheses (not the small ones) by distributing the 4
into the BIG parentheses leaving the small parentheses intact:

{{{y = 4(x+3)^2-36-13)}}}

Combine the -36-13 as -49

{{{y = 4(x+3)^2-49}}}

That's it.  Compare it to

{{{y = a(x-h)^2+k)}}}

And see that -h = +3 which means h = -3
             +k = -49 which means or k = -49

Which means the vertex point is (h,k) = (-3,-49)

and that the graph passes through the three points

1.  The vertex = (h,k) = (-3,-49)
2.  The point just to the right of the vertex: (h+1,k+a) = (-3+1,-49+4) = (-2,-45)
3.  The point just to the left of the vertex: (h-1,k+a) = (-3-1,-49+4) = (-4,-45)

Plot those three points and sketch the graph:

{{{drawing(9600/53,800,-9,3,-50,3, graph(9600/53,800,-9,3,-50,3,4x^2+24x-13),


circle(-3,-49,0.15),circle(-3,-49,0.13),circle(-3,-49,0.11),circle(-3,-49,0.09),circle(-3,-49,0.07),circle(-3,-49,0.05),circle(-3,-49,0.03),circle(-3,-49,0.01),
circle(-2,-45,0.15),circle(-2,-45,0.13),circle(-2,-45,0.11),circle(-2,-45,0.09),circle(-2,-45,0.07),circle(-2,-45,0.05),circle(-2,-45,0.03),circle(-2,-45,0.01),
circle(-4,-45,0.15),circle(-4,-45,0.13),circle(-4,-45,0.11),circle(-4,-45,0.09),circle(-4,-45,0.07),circle(-4,-45,0.05),circle(-4,-45,0.03),circle(-4,-45,0.01)
 

)}}}

Edwin</pre>