Question 538349
<pre>
The vertex form for the equation of a parabola with vertical axis of
symmetry is
 
y = a(x - h)² + k
 
Be sure to memorize that equation, and the following facts about it:
 
Its vertex is the point (h,k)
 
Its axis of symmetry is the vertical line that passes through the vertex,
which has the equation x = h
 
It passes through the two points (h+1,k+a) and (h-1,k+a)
 
If a is positive, the parabola opens upward and its vertex is a minimum point.
If a is negative, the parabola opens downward and its vertex is a maximum point.
 
The y-intercept is found by substituting 0 for x and solving for y, It
will be the point (0,"that y-value")
 
The x-intercepts, (if any), are found by substituting 0 for y and solving
for x.  They will be the points (r<sub>1</sub>, 0) and (r<sub>2</sub>, 0)
 
The minimum value is the number k if the vertex is a minimum point.
The maximum value is the number k if the vertex is a maximum point.
 
Its standard form is y = ax² + bx + c which is gotten from the vertex
form by multiplying it out, collecting like terms and placing it in
descending order.
 
Its factored form is y = a(x - r<sub>1</sub>)(x - r<sub>2</sub>) where
the r's stand for the x-values of the x-intercepts, if any.  Not all 
equations of parabolas have x-intercepts or a factored form.
 

------------------------
 
Your equation:
 
y = {{{3/4}}}(x - 2)² - 3
 
is already in vertex form.
 
We compare it to
 
y = a(x - h)² + k
 
We see that a = {{{3/4}}}, h = 2, and k = -3
 
So using the above facts:
 
Its vertex is the point (h,k) = (2,-3)
 
We plot that vertex point:
 
{{{drawing(400,400,-6,10,-6,10, 
 
graph(400,400,-6,10,-6,10), circle(2,-3,.1), locate(2,-3,"(2,3}") )}}}
 

Its axis of symmetry is the vertical line that passes through the vertex,
which has the equation x = h or x = 2
 
So through that vertex we draw the axis of symmetry. 
 
{{{drawing(400,400,-6,10,-6,10, 
 
graph(400,400,-6,10,-6,10), circle(2,-3,.1), locate(2,-3,"(2,3}"),
green(line(2,-15,2,15)) )}}}
 
It passes through the two points (h+1,k+a) and (h-1,k+a).
 
h+1 = 2+1 = 3, k+a = -3+{{{3/4}}} = -2.25, the point (2,-2.25)
h-1 = 2-1 = 1, the point (1,-2.25)
 
We plot those two points
{{{drawing(400,400,-6,10,-6,10,
graph(400,400,-6,10,-6,10), circle(2,-3,.1), locate(2,-3,"(2,3}"),
green(line(2,-15,2,15)),circle(1,-2.25,.1),circle(3,-2.25,.1)   )}}}
 
Since a is positive, the parabola opens upward and its vertex is a minimum point.
 
The y-intercept is found by substituting 0 for x and solving for y.
 
y = {{{3/4}}}(x - 2)² - 3
y = {{{3/4}}}(0 - 2)² - 3
y = {{{3/4}}}(-2)² - 3
y = {{{3/4}}}(4) - 3
y = 3 - 3
y = 0
 
So the y-intercept is the point (0,0), the origin.  We plot that
and also another matching point on the other side of the axis of
symmetry, which is (4,0), We plot those:
 
{{{drawing(400,400,-6,10,-6,10,
graph(400,400,-6,10,-6,10), circle(2,-3,.1), locate(2,-3,"(2,3}"),
green(line(2,-15,2,15)),circle(1,-2.25,.1),circle(3,-2.25,.1),
circle(0,0,.15), circle(4,0,.15) )}}}
 
Now we can draw in the parabola:

{{{drawing(400,400,-6,10,-6,10,
graph(400,400,-6,10,-6,10,

(3/4)(x - 2)^2 - 3



), circle(2,-3,.1), locate(2,-3,"(2,3}"),
green(line(2,-15,2,15)),circle(1,-2.25,.1),circle(3,-2.25,.1),
circle(0,0,.15), circle(4,0,.15) )}}}



 As it turns out in this particular problem, we have already found the
x intercepts,  (0,0) and (4,0). In other problems we would have to
find them by setting y = 0.  But we didn't need to here.
 
The minimum value is the number k or -3 because the vertex is a minimum point.
 
Its factored form is gotten from the vertex form by multiplying it out,
collecting like terms and placing it in descending order and factoring:
 
y = {{{3/4}}}(x - 2)² - 3
 
Multiply through by 4
 
4y = 3(x - 2)² - 12
4y = 3(x - 2)(x - 2) - 12
4y = 3(x² - 4x + 4) - 12
4y = 3x² - 12x + 12 - 12
4y = 3x² - 12x
4y = 3x(x - 4)

 y = {{{3/4}}}x(x - 4)
 
if you think of the x factor as (x - 0) you can see it is
factored form  y = a(x - r<sub>1</sub>)(x - r<sub>2</sub>) :

y = {{{3/4}}}(x - 0)(x - 4)

but you can leave it

 y = {{{3/4}}}x(x - 4)
 
and that's the factored form.

The standard form is found by multiplying either of the forms out and collecting terms:

If we multiply the factored form out:

 y = {{{3/4}}}x(x - 4)
 y = {{{3/4}}}(x² - 4x)
 y = {{{3/4}}}x² - 3x

That is the standard form y = ax² + bx + c if you think of it as

 y = {{{3/4}}}x² - 3x + 0

But you can just leave it as

 y = {{{3/4}}}x² - 3x

and that will be considered the standard form

Edwin</pre>