Question 84031
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What is the vertex form of y=(3x+1)(x-2)

Foil it out

y = 3x² - 6x + x - 2

y = 3x² - 5x - 2

Factor the coefficient of x², which is 3,
out of the first two terms only:  Not to 
take 3 out of -5x you divide -5 by 3 and get
{{{-5/3}}}

y = 3(x² - {{{5/3}}}x) - 2

Take one-half the coefficient of x:

{{{1/2}}}·{{{5/3}}} = {{{5/6}}}

Then square what you get:

{{{(5/6)^2}}} = {{{25/36}}}

Add that and then subtract it inside the parentheses:

y = 3(x² - {{{5/3}}}x + {{{25/36}}} - {{{25/36}}}) - 2

Change the parentheses to brackets so we can factor 
and put parentheses inside:

y = 3[x² - {{{5/3}}}x + {{{25/36}}} - {{{25/36}}}] - 2

Factor the first three terms inside the brackets:

y = 3[(x - {{{5/6}}})(x - {{{5/6}}}) - {{{25/36}}}] - 2 

Write (x - {{{5/6}}})(x - {{{5/6}}}) as (x - {{{5/6}}})²

y = 3[(x - {{{5/6}}})² - {{{25/36}}}] - 2

Now remove the brackets by distributing the 3 into the 
bracket, leaving the (x - {{{5/6}}})² intact.

y = 3(x - {{{5/6}}})² - 3·{{{25/36}}} - 2

Simplify the last two terms:

y = 3(x - {{{5/6}}})² - {{{25/12}}} - 2 

y = 3(x - {{{5/6}}})² - {{{25/12}}} - {{{24/12}}}

y = 3(x - {{{5/6}}})² - {{{49/12}}}

Compare that to:

y = a(x - h)² + k

and equate like parts of the two equations:

a = 3,

-h = {{{-5/6}}}, so h = {{{5/6}}}

k = {{{-49/12}}}

So the vertex is the point ({{{5/6}}}, {{{-49/12}}}), or

like (.8, 4.1)

or a mixed fraction is better for graphing:

The vertex is the point ({{{5/6}}}, -(4{{{1/12}}})

Nasty fractions, indeed, but nasty fractions don't bother 
computers, so why should they bother us humans?  :-)
But if we can get some more points we can plot the graph.
The graph will be a parabola and we can observe if that
point really and truly is the vertex: 

We can get the y-intercept by going back to the original
equation  y=(3x+1)(x-2) and substituting x=0. We get
y = (3·0+1)(0-2) = (0+1)(-2) = (1)(-2) = -2

So we plot those two points and a bunch of others, we get
this graph.  

{{{drawing(400,400,-1,3, -5,7,

        graph(400,400,-1,3,-5,7,(3x+1)(x-2) ),
      locate(5/6,-49/12+.3,o),
      locate( .2,-4.6,"(5/6,-4.0833),the_vertex"  ) 
       ) 
        }}}
     

Edwin</pre>