Question 413886
<pre><font face = "consolas" color = "indigo" size = 4><b>
The maximum or minimum point (vertex) of the graph of 

{{{y = ax^2+bx+c}}}

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

If a is a positive number, the vertex is a minimum.
If a is a negative number, the vertex is a maximum 

Its y-coordinate is found by substituting the value of the x
coordinate for x into {{{y = ax^2+bx+c}}}

It's zeros are 

{{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}},0) 

and

{{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}},0)

The axis of symmetry is the vertical line which passes through the 
vertex and has the equation

{{{x = -b/(2a)}}}

The y-intercept is (0,c)

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

Identify the maximum or minimum (vertex), zeros, and axis of symmetry line and show the graph.

y=x^2-4x+1

a=1, b=-4, c=1


The maximum or minimum point (vertex) of the graph of 

{{{y = ax^2+bx+c}}}

{{{y = 1x^2-4x+1}}}

a=1, b=-4, c=1

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

                 = {{{-(-4)/(2(1))}}}
  
                 = {{{4/2}}}

                 = {{{2}}}

a=1 is a positive number, so the vertex is a minimum.

Its y-coordinate is found by substituting the value of the x
coordinate for x into {{{y = ax^2+bx+c}}}

Substituting 2 for x

{{{y = 1x^2-4x+1}}}
{{{y = 1(2)^2-4(2)+1}}}
{{{y = 4-8+1}}}
{{{y = -3}}}

So the y-coordinate is -3 and so the vertex has
coordinates:

(2, -3)

It's zeros are 


({{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}},0) 

and

({{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}},0)

Substituting  a=1, b=-4, c=1


({{{(-(-4) + sqrt((-4)^2-4*1*1 ))/(2*1) }}},0)

({{{(4 + sqrt(16-4))/2 }}},0)

({{{(4 + sqrt(12))/2 }}},0)

({{{(4 + sqrt(4*3))/2 }}},0)

({{{(4 + 2sqrt(3))/2 }}},0)

({{{(2(2 + sqrt(3)))/2 }}},0)

({{{(cross(2)(2 + sqrt(3)))/cross(2) }}},0)

({{{2 + sqrt(3) }}},0)

Similarly just be changing the sign before the radical term
above, the other zero is

({{{2 - sqrt(3) }}},0)

They are approximately (.27,0) and (3.73,0)

The axis of symmetry is the vertical line which passes through the 
vertex and has the equation

{{{x = -b/(2a)}}}

which is

{{{x = 2}}}

We plot the vertex, the x-intercepts and the axis of symmetry (in green).
and the y-intercept is (0,c)

{{{drawing(400,400,-3,7,-5,5, 
graph(400,400,-3,7,-5,5), circle(2,-3,.1), locate(2,-3,"(2,-3)"),
circle(2-sqrt(3),0,.1), circle(2+sqrt(3),0,.1), green(line(2,6,2,-6)),
circle(0,1,.1) 

)}}}

Then sketch in the graph:


{{{drawing(400,400,-3,7,-5,5, 
graph(400,400,-3,7,-5,5,x^2-4x+1), circle(2,-3,.1), locate(2,-3,"(2,-3)"),
circle(2-sqrt(3),0,.1), circle(2+sqrt(3),0,.1), green(line(2,6,2,-6)),
circle(0,1,.1) 

)}}}

Edwin</pre>