

Quadratic Equations and Quadratic
Function 
Quadratic function or
the seconddegree polynomial 
Vertex
(maximum/minimum)  coordinates of translation

Roots or zeros of the
function, axis of symmetry and yintercept 
Transformations
of the quadratic function's expression 
Graphing a quadratic
function 
Transformations
of the graph of the quadratic function 





Quadratic function or
the seconddegree polynomial 
The
polynomial function of the second degree, f
(x)
=
a_{2}x^{2}
+ a_{1}x
+ a_{0}, is called a
quadratic function. 


y
= f(x)
=
a_{2}x^{2}
+ a_{1}x
+ a_{0} or
y

y_{0}
= a_{2}(x

x_{0})^{2}, 


where 

are the coordinates
of translations of the quadratic 

function. By
setting x_{0}
= 0
and
y_{0}
= 0 we obtain
y
=
a_{2}x^{2},
the source quadratic function.

The turning point V(x_{0},
y_{0})
is called the vertex of the parabola.

Note that the coefficients, a_{2}, a_{1}
and a_{0},
of quadratic function, correspond to the coefficients, a, b
and c,
of

quadratic equation, respectively.

The
real zeros of the
quadratic function: 



The above formula is known quadratic formula
that shows the
symmetry of the roots relative to the axis of

symmetry of the
parabola.


y
= f(x) =
a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{2}(x
 x_{1})(x
 x_{2})
= a_{2}[x^{2}

(x_{1} +
x_{2})x
+
x_{1}x_{2}] 


The graph of a quadratic function is curve called a parabola. The parabola is symmetric with respect to a vertical line
called the axis of symmetry. 
As
the axis of symmetry passes through the vertex of the parabola
its equation is x
= x_{0}. 
Quadratic
function has the yintercept at the
point ( 0,
a_{0 }).


Translated form of
quadratic function 
The
proof that quadratic function f
(x)
= a_{2}x^{2} + a_{1}x
+ a_{0}
is translation of its source or original
f (x)
= a_{2}x^{2} 
1)
Let calculate the
coordinates of translations of quadratic function using the
formulas, 
substitute
n
= 2 in 



then 



2)
To
get the source quadratic function we should plug the coordinates
of translations (with changed signs) 
into the general form
of the quadratic,
i.e., 

after
expanding and reducing obtained is 
y
=
a_{2}x^{2}
the source quadratic function 
3)
Inversely, by plugging the coordinates of translations into the source quadratic function 
y

y_{0}
= a_{2}(x

x_{0})^{2}, 

and
after
expanding and reducing we obtain 
y
=
a_{2}x^{2}
+ a_{1}x
+ a_{0} the quadratic function
in the general form. 

Graphing a quadratic
function 
Transformations
of the graph of the quadratic function 
How
changes in the expression of the quadratic function affect its
graph is shown in the figures below. 



The graph of quadratic polynomial will intersect the
xaxis in two
distinct points if its leading coefficient a_{2}
and the vertical translation
y_{0}
have different signs, i.e., if a_{2}
· y_{0
}<
0. 


Example:
Find zeros and vertex of
the quadratic function y
=

x^{2}
+ 2x
+
3
and sketch its graph. 
Solution:
A quadratic function can
be rewritten into translatable form y

y_{0}
= a_{2}(x

x_{0})^{2}
by completing the square, 
y
=

x^{2}
+ 2x
+ 3 
Since a_{2}
· y_{0
}<
0 given
quadratic function must have two different real zeros. 
y
=

(x^{2}

2x)
+ 3 
To find zeros of a function, we set
y
equal to zero and solve for x.
Thus, 
y
=

[(x

1)^{2}

1]
+ 3 

4
=

(x

1)^{2} 
y 
4
=

(x

1)^{2} 
(x

1)^{2}
=
4 
y

y_{0}
= a_{2}(x

x_{0})^{2} 
x

1
=
±
sqrt(4) 
V(x_{0},
y_{0})
=>
V(1,
4) 
x_{1,2}
=
1
±
2, =>
x_{1}
= 
1
and x_{2}
= 3. 


We
can deal with the given quadratic using the property of the
polynomial explored under the title, 
'
Source
or original polynomial function '. Thus, 
1)
calculate the coordinates of translations of the quadratic
y
=
f (x) =

x^{2}
+ 2x
+
3 

2)
To
get the source quadratic function, plug the coordinates
of translations (with changed signs) 
into the general form
of the quadratic, i.e., 
y
+ y_{0}
= a_{2}(x
+ x_{0})^{2}
+ a_{1}(x
+ x_{0})
+ a_{0}
=> y
+ 4
= 
(x
+ 1)^{2}
+ 2(x
+ 1)
+
3 
y
=

x^{2
}the source quadratic function 
3)
Inversely, by plugging the coordinates of translations into the source quadratic function 
y
 y_{0}
= a_{2}(x
 x_{0})^{2}
=> y
 4
= 
(x 
1)^{2}

obtained is given quadratic in general form
y
=

x^{2}
+ 2x
+
3. 








Precalculus
contents B 



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