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 $a$.
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 $x$
$f(x) = a[ x^2 + \dfrac{b}{a} x ] + c$
- 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
- $f(x) = x^2 + 2 x - 3$
- $g(x) = -x^2 + 2 x - 1$
- $h(x) = -2 x^2 + 2 x - 2$
Solution
- 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 $(-3,0)$.
- 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 $(1,0)$.
- 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 ?
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
$f(0) = 6 = a \cdot 0^2 + b \cdot 0 + c = c$
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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