SECTION 23.1 WHAT IS A
QUADRATIC SET EQUAL TO Y?
WHAT IS A QUADRATIC EXRESSION? 
An algebraic expression in the form of ax^{2} + bx + c, where a, b and c are real numbers and a ≠ 0, is called a "quadratic expression."
Please note that all the terms in the quadratic expression are ADDED. Those terms must be added in order to accurately isolate a, b and c.
Here are some examples:
ex. 1) 2x^{2} + 2x + 5
a = 2, b = 2 and c = 5
ex. 2) 3x^{2}  3x + 4
I will rewrite
3x^{2}  3x + 4 as 3x^{2} +  3x + 4
a = 3, b = 3 and c = 4
ex. 3) 3x^{2}
I will rewrite
3x^{2} as 3x^{2} + 0x + 0
a = 3, b = 0 and c = 0

Equations written in the form of y = ax^{2} + bx + c, where a, b and c are real numbers and a ≠ 0 are called "quadratics set equal to y." This type of "quadratic equation" will create ordered
pairs that produce a figure called a "parabola."
AN ASIDE: YOU HAVE BEEN FACTORING QUADRATIC EXPRESSIONS ON MANY OCCASIONS!
HOW TO FIND THE VERTEX OF A PARABOLA 
X_{vertex}= b/(2a)
Y_{vertex}: substitute the X_{vertex} into the x values of the equation of the parabola. 
Here are some examples of the above ideas:
ex. 1) Name a, b, and c, the vertex, and 4 ordered pairs of the parabola y = x^{2} + 1x + 1 and then graph the ordered pairs and the vertex.
Let's get the vertex:
a is the coefficient of x^{2} therefore a=1.
b is the coefficient of x:
therefore b=1.
c is the numerical part, i.e. the isolated number with no variable part:
therefore c = 1..
,
Let's get some ordered pairs and then create a graph:
ex. 2) Find a,b, and c; the vertex and 4 ordered pairs of the parabola y = 2x^{2} + 1;
graph the ordered pairs.
I can rewrite y =
2x^{2} + 1 as y =
2x^{2} + 0x + 1 so I can easily determine a, b, and c.
Let's get the vertex:
a is the coefficient of x^{2} : therefore a = 2.
b is the coefficient of x:
therefore b=0 since we don't have a term with x.
c is the numerical part, i.e. the isolated number with no variable part:
therefore c = 1.
Here are the ordered pairs and the graph of the parabola:
SECTION 23.2 QUADRATICS AND X
AND Y INTERCEPTS
In an earlier chapter, chapter 14 , I reintroduced the concept of x and y intercepts. I will look
at the general rules for determining the x and y intercepts of a parabola.
ex.) Given y = x^{2} + x  6
a) Find the x and y intercept
b) Find the vertex
c) Plot the vertex and the x and y intercepts.
ANSWERS:
a) To find the x intercept set y = 0.
x + 3 = 0 or x  2 = 0
3 and 2 are the x intercepts.
The associated y values are 0.
The associated points are (3, 0) and (2, 0)
To find the y intercept set x = 0.
6 is the y intercept and its associated x value is 0.
The associated point is (0, 6)
b) y= x + x  6 = 1x^{2} + 1x + 6
a = 1: b = 1 ; c = 6
c)
Also note that since a=1 is positive the graph is concave up.
I will consider an example of a parabolic equation that produces a
parabola with no x intercepts:
ex. Determine the xintercepts for y = x^{2}  4x + 9
answer:
y = 1x^{2} + 4x + 9
I rewrote the original quadratic as addition so I could get a, b, and c.
Therefore a = 1; b = 4; c = 9
To find the x  intercepts, set y = 0, then solve for x: 0 = x^{2}  4x + 9
We will use the Quadratic Formula to solve this since we can't factor x^{2}  4x + 9.
Notice that the radicand is 20 and you can't find the square root of 20.
What this reveals is:

IF YOU GET A RADICAND THAT IS NEGATIVE THEN
YOU CAN CONCLUDE THERE ARE NO XINTERCEPTS

ex.) Graph y = x^{2}  4x + 9 by getting the vertex and the y intercept and sketch the graph to see if there are no xintercepts.
Also note that since a = 1 is positive the graph is concave up.
a = 1; b = 4 ; c = 9
The vertex is: (2, 5)
To get the yintercept: set x = 0 and solve for y:
Here's the graph:
example) Graph y = 3x^{2}  14x  8 using the vertex and the x and y intercepts.
Note that since a = 3 is
negative the graph will be concave down.
Note that I rewrote the above equation in terms of addition which is the form
of the quadratic equation.
The vertex is:
yintercept:
xintercept:
or x + 4 = 0
or x = 4
HERE'S THE GRAPH:
