Quadratic Functions(General Form)
Quadratic functions and the properties of their graphs such as vertex and x and y intercepts are explored interactively using an html5 applet.
You can also use this applet to explore the relationship between the x intercepts of the graph of a quadratic function f(x) and the solutions of the corresponding quadratic equation f(x) = 0. The exploration is carried by changing values of 3 coefficients a, b and c included in the definition of f(x).
Once you finish the present tutorial, you may want to go through tutorials on quadratic functions , graphing quadratic functions and Solver to Analyze and Graph a Quadratic Function
If needed, Free graph paper is available.
A  Definition of a quadratic function
A quadratic functionf is a function of the form
f(x) = ax^{ 2} + bx + c
where a , b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a .
Examples of quadratic functions
 f(x) = 2x^{ 2} + x  1
 f(x) = x^{ 2} + 3x + 2
Interactive Tutorial (1)
Explore quadratic functions interactively using an html5 applet shown below; press "draw' button to start

Use the boxes on the left panel of the applet window to set coefficients a, b and c to the values in the examples above, 'draw' and observe the graph obtained. Note that the graph corresponding to part a) is a parabola opening down since coefficient a is negative and the graph corresponding to part b) is a parabola opening up since coefficient a is positive. You may change the values of coefficient a, b and c and observe the graphs obtained.
Answers
B  Standard form of a quadratic function and vertex
Any quadratic function can be written in the standard form
f(x) = a(x  h)^{ 2} + k
where h and k are given in terms of coefficients a , b and c .
Let us start with the quadratic function in general form and complete the square to rewrite it in standard form.

Given function f(x)
f(x) = ax^{ 2} + bx + c

factor coefficient a out of the terms in x^{ 2} and x
f(x) = a ( x^{ 2} + (b / a) x ) + c

add and subtract (b / 2a)^{ 2} inside the parentheses
f(x) = a ( x^{ 2} + (b/a) x + (b/2a)^{ 2}  (b/2a)^{ 2} ) + c

Note that
x^{ 2} + (b/a) x + (b/2a)^{ 2}

can be written as
(x + (b/2a))^{ 2}

We now write f as follows
f(x) = a ( x + (b / 2a) ) ^{ 2}  a(b / 2a)^{ 2} + c

which can be written as
f(x) = a ( x + (b / 2a) ) ^{ 2}  (b^{ 2} / 4a) + c

This is the standard form of a quadratic function with
h =  b / 2a
k = c  b^{ 2} / 4a
When you graph a quadratic function , the graph will either have a maximum or a minimum point called the vertex. The x and y coordinates of the vertex are given by h and k respectively.
Example : Write the quadratic functionf given by f(x) = 2 x^{ 2} + 4 x + 1 in standard form and find the vertex of the graph.
Solution

given function
f(x) = 2 x^{ 2} + 4x + 1

factor 2 out
f(x) = 2(x^{ 2}  2 x) + 1

We now divide the coefficient of x which is 2 by 2 and that gives 1.
f(x) = 2(x^{ 2}  2x + (1)^{ 2}  (1)^{ 2}) + 1

add and subtract (1)^{ 2} within the parentheses
f(x) = 2(x^{ 2}  2x + (1)^{ 2}) + 2 + 1

group like terms and write in standard form
f(x) = 2(x  1)^{ 2} + 3

The above gives h = 1 and k = 3.

h and k can also be found using the formulas for h and k obtained above.
h =  b / 2a =  4 / (2(2)) = 1
k = c  b^{ 2} / 4a = 1  4^{ 2}/(4(2))= 3

The vertex of the graph is at (1,3).
Interactive Tutorial (2)

Go back to the applet window and set a to 2, b to 4 and c to 1 (values used in the above example). Check that the graph opens down (a < 0) and that the vertex is at the point (1,3) and is a maximum point.

Use the applet window and set a to 1, b to 2 and c to 0, f(x) = x^{ 2}  2 x. Check that the graph opens up (a > 0) and that the vertex is at the point (1,1) and is a minimum point.
C  x intercepts of the graph of a quadratic function
The x intercepts of the graph of a quadratic function f given by
f(x) = a x^{ 2} + b x + c
are the real solutions, if they exist, of the quadratic equation
a x^{ 2} + b x + c = 0
The above equation has two real solutions and therefore the graph has x intercepts when the discriminant D = b^2  4 a c is positive. It has one repeated solution when D is equal to zero. The solutions are given by the quadratic formulas
x_{ 1} = (b + √ D)/(2 a)
and
x_{ 2} = (b  √ D)/(2 a)
Example: Find the x intercepts for the graph of each function given below
 f(x) = x_{ 2} + 2 x  3
 g(x) = x_{ 2} + 2 x  1
 h(x) = 2_{ 2} + 2 x  2
Solution

To find the x intercepts, we solve
x^{ 2} + 2 x  3 = 0
discriminant D = 2^{ 2}  4 (1)(3) = 16
two real solutions:
x_{1} = (2 + √16) / (2 * 1) = 1
and
x_{2} = (2  √16) / (2 * 1) = 3
The graph of function in part a) has two x intercepts are at the points (1,0) and (3,0).

We solve x^{ 2} + 2 x  1 = 0
discriminant D = 2^{ 2}  4(1)(1) = 0
one repeated real solutions x_1 = b / 2a = 2 / 2 = 1
The graph of function in part b) has one x intercept at (1,0).

We solve 2 x^{ 2} + 2 x  2 = 0
discriminant D = 2^{ 2}  4(2)(2) = 12
No real solutions for the above equation
No x intercept for the graph of function in part c).
Interactive Tutorial (3)

Go to the applet window and set the values of a, b and c for each of the examples in parts a, b and c above and check the discriminant and the x intercepts of the corresponding graphs.

Use the applet window to find any x intercepts for the following quadratic functions.
a) f(x) = x^{ 2} + x  2
b) g(x) = 4 x^{ 2} + x + 1
a) h(x) = x^{ 2}  4 x + 4
Use the analytical method described in the above example to find the x intercepts and compare the results.

Use the applet window and set a, b and c to values such that b^{ 2}  4 a c < 0.
How many xintercepts does the graph of f(x) have ?

Use the applet window and set a, b and c to values such that b^{ 2}  4 a c = 0. How many xintercepts the does the graph of f(x) have?

Use the applet window and set a, b and c to values such that b^{ 2}  4ac > 0.
How many xintercepts does the graph of f(x) have ?
Answers
D  y intercepts of the graph of a quadratic function
The y intercept of the graph of a quadratic function is given by f(0) = c .
Example: Find the y intercept of the graph of the following quadratic functions.
 f(x) = x^{ 2} + 2 x  3
 g(x) = 4 x^{ 2}  x + 1
 h(x) = x^{ 2} + 4 x + 4
Solution

f(0) = 3. The graph of f has a y intercept at (0,3).

g(0) = 1. The graph of g has a y intercept at (0,1).

h(0) = 4. The graph of h has a y intercept at (0,4).
Interactive Tutorial (4)

Use the applet window to check the y intercept for the quadratic functions in the above example.

Use the applet window to check the y intercept is at the point (0,c) for different values of c.
As an exercise you are asked to find the equation of a quadratic function whose graph is shown in the applet and write it in the form f(x) = a x^{ 2} + b x + c.
USE this applet to Find Quadratic Function Given its Graph
Example: Find the graph of the quadratic function f whose graph is shown below.
Solution
There are several methods to answer the above question but all of them have one idea in common: you need to understand and then select the right information from the graph.
method 1:
The above graph has two x intercepts at (3,0) and (1,0) and a y intercept at (0,6) . The x coordinates of the x intercepts can be used to write the equation of function f as follows:
f(x) = a(x + 3)(x + 1)
We now use the y intercept f(0) = 6
6 = a(0 + 3)(0 + 1)
and solve for a to find a = 2 . The formula for the quadratic function f is given by :
f(x) = 2(x + 3)(x + 1) = 2 x^{ 2} + 8 x + 6
method 2:
The above parabola has a vertex at (2 , 2) and a y intercept at (0,6) . The standard (or vertex) form of a quadratic function f can be written
f(x) = a(x + 2)^{ 2}  2
We use the y intercept f(0) = 6
6 = a(0 + 2)^{ 2}  2 . Solve for a to find a = 2 . The formula for the quadratic function f is given by :
f(x) = 2(x + 2)^{ 2}  2 = 2 x^{ 2} + 8 x + 6
method 3:
Since a quadratic function has the form
f(x) = a x^{ 2} + b x + c
we need 3 points on the graph of f in order to write 3 equations and solve for a , b and c .
The following points are on the graph of f
(3 , 0) , (1 , 0) and (0 , 6)
point (0 , 6) gives
f(0) = 6 = a * 0^{ 2} + b * 0 + c = c
solve for c to obtain c = 6
The two other points give two more equations
(3 , 0) gives f(3) = a * (3)^{ 2} + b * (3) + 6
which leads to 9 a  3 b + 6 = 0
and (1 , 0) gives f(3) = a (1)^{ 2} + b * (1) + 6
which becomes a  b + 6 = 0
Solve the last two equations in a and b to obtain
a = 2 and b = 8 and gives
f(x) = 2 x^{ 2} + 8 x + 6
Go back to the applet above, generate a graph and find its equation. You can generate as many graphs, and therefore questions, as you wish.
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