AOS1 Topic 3: Quadratic Functions

A quadratic function is a type of polynomial function that can be written in the standard form:


\[ f(x) = ax^2 + bx + c \]

where:


$a$ , $ b$ , and $ c$ are constants, with \(a \neq 0\).

The highest exponent of the variable \(x\) is $2$, making it a second-degree polynomial.


Key Features of Quadratic Functions


1. Parabola Shape

The graph of a quadratic function is a U-shaped curve called a parabola . The direction in which the parabola opens depends on the coefficient $a$

  1. Upwards if \(a > 0\)
  2. Downwards if \(a < 0\)
  3. Wide if $a$ is close to $0$
  4. Skinny if $a$ is far greater or far lesser than $0$



2. Vertex


The vertex is the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards, respectively. It represents the maximum or minimum value of the quadratic function.


The vertex \((h, k)\) can be found using the formula:


\[ h = - \frac{b}{2a} \]


\[ k = c - \frac{b^2}{4a} \]


Created with GeoGebra ® , by Rabiya Manzoor, Link



3. Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror images. It passes through the vertex.


\[ x = h = -\frac{b}{2a} \]


Example:


For the quadratic function $f(x)=−3x^2+6x+2$, find the axis of symmetry.


Solution:


\[h=− \frac{b}{2a} =− \frac{6}{2×(−3)} = − \frac{6}{−6} = 1 \]


Axis of Symmetry: $x=1$



4. Y-Intercept


The y-intercept is the point where the graph crosses the y-axis. It occurs when \(x = 0\), so the y-intercept is \(f(0) = c\).


Example:


Find the y-intercept of the quadratic function $f(x)=4x^2−2x+5$.


Solution:


\[f(0)=4(0)^2−2(0)+5=5\]


Y-Intercept: $(0,5)$



5. X-Intercepts (Roots)


The x-intercepts are the points where the parabola crosses the x-axis. These can be found by solving the quadratic equation \(ax^2 + bx + c = 0\). The solutions (roots) can be found using the quadratic formula :


\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]


The discriminant, \( \Delta = b^2 - 4ac\), determines the nature of the roots:


  1. If \( \Delta > 0\), there are two real and distinct roots.
  2. If \( \Delta = 0\), there is one real, repeated root.
  3. If \( \Delta < 0\), there are no real roots (the parabola does not cross the x-axis).


Example:

Find the x-intercepts of the quadratic function $f(x)=x^2−4x+3$.


Solution:


  1. Calculate Discriminant


  1. \[ \Delta =(−4)^2−4(1)(3)=16−12=4>0 \]


  1. Since $ \Delta >0$, there are two real and distinct roots.


  1. Find the roots:


\[ x = \frac{b \pm \sqrt{b^2-4ac}}{2a} = \frac{4 \pm \sqrt{4}}{2} \]

\[ x = \frac{4 \pm 2}{2} = 2 \pm 1\]


\[x=2+1=3\]

\[x=2-1=1\]


  1. X-Intercepts: $(1,0)$ and $(3,0)$



Forms of Quadratic Functions


1. Standard Form


The general form of a quadratic function is:

\[ f(x) = ax^2 + bx + c \]


Uses:

  1. Identifying the y-intercept ($c$)
  2. Applying the quadratic formula to find roots


Example:

Convert the quadratic function $f(x)=3x^2−6x+2$ to standard form.


Solution:

The function is already in standard form: $f(x)=3x^2−6x+2$.



2. Vertex Form


The vertex form of a quadratic function is:

\[ f(x) = a(x - h)^2 + k \]


where $(h,k)$ is the vertex of the parabola.


Advantages:


  1. Easily identifies the vertex and axis of symmetry
  2. Simplifies graphing and transformations


Example:

Convert the standard form $f(x)=2x^2+8x+6$ to vertex form.


Solution:


Factor out $a=2$ from the first two terms:

\[f(x)=2(x^2+4x)+6\]


Complete the square inside the parentheses:

  1. Take half of 4, which is 2, and square it, getting 4.
  2. Add and subtract 4 inside the parentheses:


\[ f(x)=2(x^2+4x+4−4)+6\]

\[f(x)=2((x+2)^2−4)+6\]

\[f(x)=2(x+2)^2−8+6\]

\[f(x)=2(x+2)^2−2\]


Vertex Form:

\[f(x)=2(x+2)^2−2 \]


Vertex: $(−2,−2)$



3. Factored Form


The factored form of a quadratic function is:

\[ f(x) = a(x - r_1)(x - r_2) \]


where $r_1$​ and $r_2$​ are the roots (x-intercepts) of the quadratic equation.


Advantages:

  1. Directly shows the x-intercepts
  2. Simplifies finding the roots


Example:

Express the quadratic function $f(x)=x^2−5x+6$ in factored form.


Solution:


Find the roots by factoring:

\[ x^2−5x+6 = (x−2)(x−3)\]


Factored Form:

\[f(x)=(x−2)(x−3)\]


Roots: $x=2$ and $x=3$



Practise factorising with the following simulation


Created with GeoGebra ® , by Mrs. Nisbet, Link




Domain and Range of Quadratic Functions


Domain of Quadratic Function


A quadratic function is a polynomial function and is defined for all real values of $x$.


So, the domain of a quadratic function is the set of real numbers, that is, \( \mathbb{R} \). In interval notation, the domain of any quadratic function is \( (-\infty, \infty) \).


Domain: $ \mathbb{R} $


Range of Quadratic Function


The range of a quadratic function depends on the direction in which the parabola opens and the vertex's y-coordinate.


If $a>0$ (Parabola Opens Upwards):


\[ \text{Range: } y \geq k \text{ or } [k, \infty)\]


If $a<0$ (Parabola Opens Downwards):


\[ \text{Range: } y \leq k \text{ or } (- \infty, k] \]


where $(h,k)$ is the vertex of the parabola.


Example:

Determine the range of the quadratic function $f(x)=−x^2+4x−3$.


Solution:


Identify coefficients:

$a=−1<0$ (parabola opens downwards)


Find vertex $(h,k)$:


\[ h=− \frac{b}{2a}=− \frac{4}{2(−1)} = 2\]

\[ k=f(2)=−(2)^2+4(2)−3=−4+8−3=1\]


Range:

\[ y \leq 1 \text{ or } (− \infty,1]\]



Summary: Quadratic Functions


Definition: A quadratic function is a second-degree polynomial function of the form $f(x)=ax^2+bx+c$


Parabola Shape: The graph is a U-shaped curve called a parabola, opening upwards if $a>0$ and downwards if $a<0$.


Vertex: The highest or lowest point on the parabola, calculated using $h=− \frac{b}{2a}$ and $k=f(h)$.


Axis of Symmetry: A vertical line passing through the vertex, given by $x=h$.


Y-Intercept: The point where the graph crosses the y-axis, found by evaluating $f(0)=c$.


X-Intercepts (Roots): Points where the graph crosses the x-axis, found using the quadratic formula $x= \frac{-b \pm \sqrt{b^2−4ac}}{2a}$.


Forms of Quadratic Functions:

  1. Standard Form: $f(x)=ax^2+bx+c$
  2. Vertex Form: $f(x)=a(x−h)^2+k$
  3. Factored Form: $f(x)=a(x−r1)(x−r2)$


Domain and Range:


Domain: All real numbers, $ \mathbb{R}$ or $(-\infty, \infty)$.


Range: Depends on the parabola's orientation:


$y \geq k$ if $a>0$

$y \leq k$ if $a<0$

Example 1

Example 1: Find the vertex, axis of symmetry, and intercepts for the quadratic function:
\[ f(x) = 2x^2 - 4x + 1 \]

Solution:
Vertex:
Using the formula \(h = -\frac{b}{2a} = -\frac{-4}{2(2)} = 1\).
Substitute \(x = 1\) into the function to find \(k\):
\[ f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1 \]
So, the vertex is \((1, -1)\).
Axis of Symmetry:
The axis of symmetry is \(x = 1\).
Y-Intercept:
Substitute \(x = 0\) into the function:
\[ f(0) = 2(0)^2 - 4(0) + 1 = 1 \]
So, the y-intercept is \((0, 1)\).

X-Intercepts:
Use the quadratic formula to find the roots:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} \]
\[ x = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2} \]
So, the x-intercepts are approximately \(x = 1.71\) and \(x = 0.29\).




Example 2

Solve the quadratic equation using the quadratic formula
Given the quadratic equation:
\[ 3x^2 - 5x + 2 = 0 \]

Solution:
Here, \(a = 3\), \(b = -5\), and \(c = 2\). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substitute the values of \(a\), \(b\), and \(c\):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(2)}}{2(3)} \]
\[ x = \frac{5 \pm \sqrt{25 - 24}}{6} \]
\[ x = \frac{5 \pm \sqrt{1}}{6} \]
\[ x = \frac{5 \pm 1}{6} \]
So, the two solutions are:
  • \[ x = \frac{5 + 1}{6} = 1 \]
  • \[ x = \frac{5 - 1}{6} = \frac{2}{3} \]
Therefore, the solutions are \(x = 1\) and \(x = \frac{2}{3}\).


Example 3

Find the maximum value of the quadratic function
Given the quadratic function:
\[ f(x) = -2x^2 + 8x - 3 \]

Solution:
The function opens downwards because \(a = -2\). Therefore, it has a maximum value at its vertex.
First, find the vertex:
\[ h = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2 \]
Substitute \(x = 2\) into the function to find \(k\):
\[ f(2) = -2(2)^2 + 8(2) - 3 = -8 + 16 - 3 = 5 \]
So, the maximum value of the function is 5, and it occurs at \(x = 2\).


Example 4

Calculating Domain and Range of Quadratic Function:
Consider the quadratic function \( f(x) = x^2 - 4x + 3 \). What is the domain and range of this function?

Solution:
Domain
The given quadratic function is \( f(x) = x^2 - 4x + 3 \), which is a polynomial function. Polynomial functions are defined for all real numbers, so the domain of this quadratic function is all real numbers.
Therefore, the domain of \( f(x) \) is:
\( \text{Domain of } f(x): (-\infty, \infty) \)

Range
First, we need to find the vertex of the quadratic function. The standard form of the quadratic function is:
\( f(x) = ax^2 + bx + c \)
For \( f(x) = x^2 - 4x + 3 \), we have \( a = 1 \), \( b = -4 \), and \( c = 3 \).
The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula:
\( x = -\frac{b}{2a} \)
Substitute the values of \( a \) and \( b \):
\( x = -\frac{-4}{2(1)} = 2 \)
Now, substitute \( x = 2 \) into the function to find the corresponding \( y \)-value:
\( f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \)
The vertex of the parabola is \( (2, -1) \). Since the coefficient of \( x^2 \) is positive (\( a = 1 \)), the parabola opens upwards. Therefore, the range is all values greater than or equal to the \( y \)-coordinate of the vertex, which is -1.
Thus, the range of \( f(x) \) is:
\( \text{Range of } f(x): [-1, \infty) \)



Exercise 1

Exercise 2

Exercise 3

Exercise 4