Question 356772
find the vertex of the x and y coordinate, line of symmetry and maximum of f(x)
f(x)=-2x^2+2x+7.
<pre>
There are two methods, {1) completing the square using the memorized
standard form f(x) = ax²+b+c 

and 

(2) using the vertex formula that you have memorized.

-----------------------------------------

Method (1)

f(x) = -2x² + 2x + 7

Factor out -2 from the first two terms:

f(x) = -2(x² - x) + 7

To the side multiply the coefficient of x, which is -1, by {{{1/2}}} getting
{{{-1/2}}} then squaring {{{-1/2}}} getting {{{(-1/2)^2}}} or {{{1/4}}}.  Then
adding that and subtracting that inside of the parentheses:

f(x) = -2(x² - x + {{{1/4}}} - {{{1/4}}}) + 7

Change the parentheses to brackets (so you can put parentheses inside):

f(x) = -2[x² - x + {{{1/4}}} - {{{1/4}}}] + 7

Factor the first three terms inside the brackets:

f(x) = -2[(x - {{{1/2}}})(x - {{{1/2}}}) - {{{1/4}}}] + 7

Since those factors in parentheses are the same we write them
as a perfect square:

f(x) = -2[(x - {{{1/2}}})² - {{{1/4}}}] + 7
 
Now remove the brackets by using the distributive
principle.  That is, we multiply the -2 by putting
it in front of the (x - {{{1/4}}})^2 and we multiply
the -2 also by the {{{-1/2}}} getting -1. So we have

f(x) = -2(x - {{{1/2}}})² + {{{1/2}}} + 7

Then we combine the two terms on the right side and get

f(x) = -2(x - {{{1/2}}})² + {{{1/2}}} + {{{14/2}}}

f(x) = -2(x - {{{1/2}}})² + {{{15/2}}}
  
We recognize this as in the standard form we have memorized:

f(x) = a(x - h)² + k

we know that a = -2, h = {{{1/2}}} and k = {{{15/2}}}

So that the vertex is (h,k) or ({{{1/2}}}, {{{15/2}}}) 

----------------

Method (2), using the vertex formula we have memorized:

x-coordinate of vertex = {{{-b/(2a)}}}

y-coordinate of vertex = what you get when you substitute the
x-coordinate for x in the equation and simplify.

f(x) = -2x² + 2x + 7

Compare to the general form we have memorized,

f(x) = ax² + bx + c

a = -2, b = 2, c = 7

x-coordinate of vertex = {{{-b/(2a)}}} = {{{-(2)/(2(-2))}}} = 
{{{(-2)/(-4)}}} = {{{1/2}}}

y-coordinate of vertex = what we get when you substitute the
x-coordinate, {{{1/2}}} for x in the equation and simplify:

f({{{1/2}}}) = -2({{{1/2}}}² + 2{{{1/2}}} + 7 = -2({{{1/4}}}) + 1 + 7 = {{{-1/2}}} + 8 = {{{-1/2}}} + {{{16/2}}} = {{{15/2}}}

so the vertex is the point V({{{1/2}}}, {{{15/2}}}).

-------

No that we have found the vertex by either of the above two methods,

we plot that point and draw a verticle line, the line of symmetry,
through it, like this line drawn in green:

{{{drawing(400,1600/3,-4,8,-6,10,
graph(400,1600/3,-4,8,-6,10),  

green(line(1/2,11,1/2,-7)),

line(1/2+.1,15/2,1/2-.1,15/2),line(1/2,15/2+.1,1/2,15/2-.1),line(1/2+.1,15/2+.1,1/2-.1,15/2-.1),line(1/2+.1,15/2-.1,1/2-.1,15/2+.1),
locate(1/2+.3,15/2+.5,V(1/2,15/2))  )}}}

That green line of symmetry has the equation x = {{{1/2}}} because
every point on that green line of symmetry has {{{1/2}}} as its 
x-coordinate.

Now we can get some other points on that graph:

 x| y
-----
-2|-5 
-1| 3
 0| 7
 1| 7
 2| 3

{{{drawing(400,1600/3,-4,8,-6,10,
graph(400,1600/3,-4,8,-6,10),

green(line(1/2,11,1/2,-7)),

line(1/2+.1,15/2,1/2-.1,15/2),line(1/2,15/2+.1,1/2,15/2-.1),line(1/2+.1,15/2+.1,1/2-.1,15/2-.1),line(1/2+.1,15/2-.1,1/2-.1,15/2+.1),
locate(1/2+.3,15/2+.5,V(1/2,15/2)),

line(-2+.1,-5,-2-.1,-5),line(-2,-5+.1,-2,-5-.1),line(-2+.1,-5+.1,-2-.1,-5-.1),line(-2+.1,-5-.1,-2-.1,-5+.1),locate(-3-.7,-5+.3,"(-2,-5)"),

line(-1+.1,3,-1-.1,3),line(-1,3+.1,-1,3-.1),line(-1+.1,3+.1,-1-.1,3-.1),line(-1+.1,3-.1,-1-.1,3+.1),locate(-2-.5,3+.2,"(-1,3)"),

line(2+.1,3,2-.1,3),line(2,3+.1,2,3-.1),line(2+.1,3+.1,2-.1,3-.1),line(2+.1,3-.1,2-.1,3+.1),locate(2-.5,3+.2,"(2,3)"),



line(0+.1,7,0-.1,7),line(0,7+.1,0,7-.1),line(0+.1,7+.1,0-.1,7-.1),line(0+.1,7-.1,0-.1,7+.1),locate(-2+.3,7+.3,"(0,7)"),

line(1+.1,7,1-.1,7),line(1,7+.1,1,7-.1),line(1+.1,7+.1,1-.1,7-.1),line(1+.1,7-.1,1-.1,7+.1),locate(1+.4,7,"(1,7)")



  )}}}

And the graph is

{{{drawing(400,1600/3,-4,8,-6,10,
graph(400,1600/3,-4,8,-6,10,-2x^2+2x+7),

green(line(1/2,11,1/2,-7)),

line(1/2+.1,15/2,1/2-.1,15/2),line(1/2,15/2+.1,1/2,15/2-.1),line(1/2+.1,15/2+.1,1/2-.1,15/2-.1),line(1/2+.1,15/2-.1,1/2-.1,15/2+.1),
locate(1/2+.3,15/2+.5,V(1/2,15/2)),

line(-2+.1,-5,-2-.1,-5),line(-2,-5+.1,-2,-5-.1),line(-2+.1,-5+.1,-2-.1,-5-.1),line(-2+.1,-5-.1,-2-.1,-5+.1),locate(-3-.7,-5+.3,"(-2,-5)"),

line(-1+.1,3,-1-.1,3),line(-1,3+.1,-1,3-.1),line(-1+.1,3+.1,-1-.1,3-.1),line(-1+.1,3-.1,-1-.1,3+.1),locate(-2-.5,3+.2,"(-1,3)"),

line(2+.1,3,2-.1,3),line(2,3+.1,2,3-.1),line(2+.1,3+.1,2-.1,3-.1),line(2+.1,3-.1,2-.1,3+.1),locate(2-.5,3+.2,"(2,3)"),



line(0+.1,7,0-.1,7),line(0,7+.1,0,7-.1),line(0+.1,7+.1,0-.1,7-.1),line(0+.1,7-.1,0-.1,7+.1),locate(-2+.3,7+.3,"(0,7)"),

line(1+.1,7,1-.1,7),line(1,7+.1,1,7-.1),line(1+.1,7+.1,1-.1,7-.1),line(1+.1,7-.1,1-.1,7+.1),locate(1+.4,7,"(1,7)")



  )}}}

The maximum value is the greatest y-value on the graph, which is the
y-coordinate of the vertex, {{{15/2}}}


Edwin</pre>