@ Solution of Quadratic
Function with Rational Roots @
A Quadratic
Function is an Algebraic equation of two variables. One variable
is a second
degree while the other is first
then usually followed by a constant.
Graph or curve (Set of Pts) of a Quadratic Function resembles a U shape
curve.
One variable is called the independent variable while the other is called the
dependent variable. The independent is the horizontal variable while the
dependent variable is the vertical variable.
Use
Gizmo to explore Quadratics!
Quadratic Functions (equations) normally appear in distinct
arrangements:
Standard
Form: f(x) = ax2 + bx + c, A,B,C are real numbers.
Other essential characteristics is that (a) is nonzero real number and f(x)
= y
also, (x) is
considered the Domain
(H values) while (y) the Range (V values).
If
the Domain of X is real numbers then a Quadratic Curve will cross
the X axis twice.
Locate: Intercepts, Axis of Symmetry, Vertex,
Symmetry Point to sketch solution!
Given: y= –x2
+x –6 Determine Infinite Solution Set which
is a U shaped curve!
What do
your suspect the Infinite Solution Set will look like?
Try to factor the Quadratic first to determine if it has Rational Roots or X intercepts!
Given: y = –x2
–x + 6 = (–x + 2) (x +
3) are factors thus x = –3 & x = +2
Rational Roots!
The (X)
Intercepts are – 3 and +2 while (Y) Intercept is y = (0)2 –(0)
+6 = +6 (0,6)
Using
the Midpoint Formula: ( X1 + X2) /
2 = + 2 + – 3 = –
1/2 Axis of Symmetry!
Substitute
Midpoint to locate Vertex: y = (–1/2)2
– (– 1/2) – 6 = +1/4 + 1/2 + 6 = + 6 3/4
Thus: Vertex = (–
1/2 , + 6 1/4 ) Locate Symmetry Point by using
Axis of Symmetry!
Using
Property of Symmetry and (Y) Intercept locate the Symmetry Point! Why?
Symmetry
Point: y = x2 –x +6 = (– 1)2 – (+ 1) + 6 =
+ (1) – (1) + 6 = + 6 ((0, + 6)!
Infinite
Solution Set of y = –x2 –x + 6 equals Graph of y = –x2 –x + 6 shown below!
|
X |
-3 |
+2 |
0 |
–1 |
– 1/2 |
|
Y |
0 |
0 |
+ 6 |
+ 6 |
+ 6 1 /4 |
Tom Love