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 and graphing quadratic functions.

If needed, Free graph paper is available.

## A - Definition of a quadratic function

A quadratic function $f$ 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 $a$.

1. $f(x) = -2x^2 + x - 1$
2. $f(x) = x^2 + 3x + 2$

Interactive Tutorial (1)

Explore quadratic functions interactively using an html5 applet shown below; press "draw' button to start

 a = 1 -10+10 b = 0 -10+10 c = 1 -10+10
>

• 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.

## 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 $x$
$f(x) = a[ x^2 + \dfrac{b}{a} x ] + c$

• add and subtract $(\dfrac{b}{2 a})^2$ inside the parentheses
$f(x) = a[ x^2 + \dfrac{b}{a} x + (\dfrac{b}{2a})^2 - (\dfrac{b}{2a})^2 ] + c$

• Note that
$x^2 + \dfrac{b}{a}x + (\dfrac{b}{2a})^2$

• can be written as
$[x + (\dfrac{b}{2a})]^2$

• We now write $f$ as follows
$f(x) = a[ x + (\dfrac{b}{2a}) ]^2 - a(\dfrac{b}{2a})^2 + c$

• which can be written as
$f(x) = a[ x + (\dfrac{b}{2a}) ]^2 - (\dfrac{b^2}{4 a}) + c$

• This is the standard form of a quadratic function with
$h = -\dfrac{b}{2 a}$

$k = c - \dfrac{b^2}{4 a}$

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 function $f$ 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.
$h = -\dfrac{b}{2 a} = -\dfrac{4}{2 \cdot (-2)} = 1$

$k = c - \dfrac{b^2}{4 a} = 1 - \dfrac{4^2}{4 \cdot (-2)}= 3$

• 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 ($a \lt 0$) 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 ($a \gt 0$) 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

$x_1 = \dfrac{-b + \sqrt D}{2 a}$

and
$x_2 = \dfrac{-b - \sqrt D}{2 a}$

Example: Find the x intercepts for the graph of each function given below

1. $f(x) = x^2 + 2 x - 3$
2. $g(x) = -x^2 + 2 x - 1$
3. $h(x) = -2 x^2 + 2 x - 2$

Solution
1. To find the $x$ intercepts, we solve

$x^2 + 2 x - 3 = 0$

discriminant $D = 2^2 - 4 \cdot 1 \cdot (-3) = 16$

two real solutions:
$x_1 = \dfrac{-2 + \sqrt {16}}{2 \cdot 1} = 1$
and
$x_2 = \dfrac{-2 - \sqrt {16}}{2 \cdot 1} = -3$

The graph of function in part a) has two $x$ intercepts are at the points $(1,0)$ and $(-3,0)$.

2. We solve $-x^2 + 2 x - 1 = 0$

discriminant $D = 2^2 - 4 \cdot (-1) \cdot (-1) = 0$

one repeated real solutions $x_1 = -\dfrac{b}{2a} = \dfrac{-2}{-2} = 1$

The graph of function in part b) has one x intercept at $(1,0)$.

3. We solve $-2 x^2 + 2 x - 2 = 0$

discriminant $D = 2^2 - 4 \cdot (-2) \cdot (-2) = -12$

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 $b^2 - 4 a c \lt 0$. How many $x$-intercepts 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 x-intercepts the does the graph of f(x) have?
• Use the applet window and set $a$, $b$ and $c$ to values such that $b^2 - 4ac \gt 0$. How many $x$-intercepts does the graph of f(x) have ?

## 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.
1. $f(x) = x^2 + 2 x - 3$
2. $g(x) = 4 x^2 - x + 1$
3. $h(x) = -x^2 + 4 x + 4$
Solution
1. $f(0) = -3$. The graph of $f$ has a $y$ intercept at $(0,-3)$.
2. $g(0) = 1$. The graph of g has a $y$ intercept at $(0,1)$.
3. $h(0) = 4$. 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 $c$.

## E - Exercises: Find the equation of a quadratic function given its graph

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

$f(0) = 6 = a \cdot 0^2 + b \cdot 0 + c = c$
solve for $c$ to obtain $c = 6$
The two other points gives two more equations

$(-3 , 0)$ gives $f(-3) = a \cdot (-3)^2 + b \cdot (-3) + 6$

which leads to $9 a - 3 b + 6 = 0$

and $(-1 , 0)$ gives $f(-3) = a (-1)^2 + b \cdot (-1) + 6$

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 graph, therefore question, as you wish.

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Updated: 2 April 2013