# Quadratic Functions

Author

bala rajakumar

Last Updated

před 8 lety

License

Creative Commons CC BY 4.0

Abstract

Notes on quadratic functions for Mathematics Standard Level

(Hey Mr.Zamar)

Author

bala rajakumar

Last Updated

před 8 lety

License

Creative Commons CC BY 4.0

Abstract

Notes on quadratic functions for Mathematics Standard Level

(Hey Mr.Zamar)

```
\documentclass[11pt]{article}
\usepackage{graphicx}
\begin{document}
\title{Quadratic Functions}
\author{Bala Rajakumar and Andi Zhang \\
Mathematics Standard Level}
\maketitle
\section{Introduction}
Quadratic Functions
\subsection{What are they?}
Quadratic Functions are functions in the form of $f(x)=ax^2+bx+c$. Where variables a, b, nor c can equal to $0$.
\\
Here are some examples of Quadratic Function that we will look at:
$1. f(x)=x^2-7x+12 \\$
$2. f(x)= 4x^2+20x+25 \\$
$3. f(x)=-3x^2-5x+7$
\subsection{How do we solve these Quadratic Functions?}
There are many ways to solve Quadratic Functions but we will only focus on two famous methods: \\
$1.$ \textbf{Quadratic Formula} \\
$2.$ \textbf{Graphing Method}
\section{Quadratic Equation}
\subsection{What is a Quadratic Equation?}
Quadratic Equation is an equation used to solve any Quadratic Function. We can follow just a formula to get the answer for x. \\
\textbf{The formula} \\
$\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}$
Using this formula allows us to get the answer quickly.
\subsection{How do we use it?}
A great Question. Well we basically see the Quadratic Function \\ $(ax^2+bx+c)$.\\
Then we just match the letters with the quadratic formula, so we place the \textbf{A} from $\underline{A}x^2$ to the \underline{a} from the Quadratic Equation. The same rule applies to \textbf{B} from $\underline{B}x$ and \textbf{C} from $\underline{c}$.
\subsection{How can we be sure this formula works?}
Here is the proof that supports this formula:
$ax^2+bx+c=0$ \textbf{$-$ Step $1$ }
$ax^2+bx=-c$ \textbf{$-$ Step 2}
$x^2+\frac{bx}{a}=\frac{-c}{a}$ \textbf{$-$ Step 3}
$x^2+\frac{bx}{a}+\frac{b}{2a}^2=\frac{-c}{a}+\frac{b}{2a}^2$ \textbf{$-$ Step 4}
$(x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{2a^2}$ \textbf{$-$ Step 5}
$x+\frac{b}{2a}=\pm \sqrt \frac{{b^2 - 4ac}} {4a^2} $ \textbf{$-$ Step6}
$x+\frac{b}{2a}= \frac{\sqrt{b^2-4ac}}{2a}$ \textbf{$-$ Step 7}
$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}$ \textbf{$-$ Step 8}
This helps us understand how the quadratic equation was formed thus gives us a good insight to trusting this formula.
\subsection{Lets solve :D}
$1. f(x)=x^2-7x+12$ \\
\underline{Step 1. Lets identify the \textbf{A,B, and C}}
In this case $x^2$ does not have a coefficient in front of it, so therefore there must be a $1$ beside $x^2$. Thus the value for \textbf{A} is going to be $1$. \\
Then the coefficient beside $x$ is $-7$, thus the value for \textbf{B} becomes $-7$
Finally C, C is going to be $12$.\\
$A=1, B=-7, C=12$ \\
\underline{Step 2. Time to use the quadratic Formula \\}
$$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}$$ \\
So we can plug in the variables \\
$$ x = \frac{{ -(-7) \pm \sqrt {(-7)^2 - 4(1)(12)} }}{{2(1)}}$$ \\
$$ x = \frac{{ 7 \pm \sqrt {1} }}{{2}}$$ \\
$$ x = \frac{{ 7 \pm 1 }}{{2}}$$ \\
Therefore $x= 4/3$ This is supported by the fact that:
$ x= \frac{8}{2}$=$4$ which was created by $ x= \frac{(7+1)}{2}$ \\
and also \\
$x= \frac{6}{2}$=$3$ which was created by $ x= \frac{(7-1)}{2}$
\includegraphics{line.png}
$ 2.f(x)= 4x^2+20x+25$ \\
\underline{Step 1. Lets identify the \textbf{A,B, and C}} \\
In this case, $\underline{4}x^2$, the $4$ is going to be the variable \textbf{A}.
Variable B is going to be $20$ since $\underline{20}x$ has the coefficient is 20.
Variable c is going to be 25.\\
$A=4, B=20, C= 25$ \\
\underline{Step 2. Using the formula} \\
$$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}$$ \\
Now we just plug in the values: \\
$$x = \frac{{ - (20) \pm \sqrt {(20)^2 - 4(4)(25)} }}{{2(4)}}$$ \\
$$x = \frac{{ - 20 \pm \sqrt {0} }}{{8}}$$ \\
$$x = \frac{{ - 20 \pm 0 }}{{8}}$$ \\
Therefore $x= -2.5$ \\
Since $\frac{-20}{8}$= $-2.5$ which was got by $\frac{-20+0}{8}$ and $\frac{-20-0}{8}$. Both these equations equal the same answer. \\
\includegraphics{line.png}
$$3. f(x)=-3x^2-5x+7$$
\underline{Step $1.$ Find the variable \textbf{A,B,C}}
$$3. f(x)=-3x^2-5x+7$$
So Variable \textbf{A} is going to be $-3$ since $-3$ is in $\underline{-3}x^2$.
Variable is B is going to be $-5$ since $-5$ is in $\underline{-5}x$. Finally, Variable \textbf{C}
is going to be $7$. \\
$A=-3, B=-5, C=7$ \\
\underline{Step $2$. Using the formula :D}
$$x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}$$ \\
Plug in the variables:
$$x = \frac{{ - (-5)\pm \sqrt {(-5)^2 - 4(-3)(7)} }}{{2(-3)}}$$ \\
$$x = \frac{{ 5\pm \sqrt {109} }}{{-6}}$$ \\
$$x = \frac{{ 5\pm 10.4 }}{{-6}}$$ \\
Therefore $x=-2.5$ or $.9$
We got this from doing this:
$$x = \frac{{ 15.4 }}{{-6}}=-2.5$$ \\
$$x = \frac{{-5.4 }}{{-6}}= 0.9$$
\section{What can we do with Quadratic Functions?}
Quadratic Functions are really useful as they create parabolas when graphed.
\subsection{Parabola's}
A parabola is U shaped figure that forms when graphed by a quadratic function. The most common Quadratic Function that draws a parabola would be $y=x^2$. \\
\includegraphics{image18-300x300.jpg}
\subsubsection{So what's so good about parabola's}
Parabola's play an important part in our architecture lives, an example of this would be bridges. Bridges, itself is a parabola, we need bridges to be a parabola because parabola's are the most strongest shape that can hold lots of cars.\\
Here are some other examples of parabola's: \\
$1$. Car headlights \\
$2$. Heaters \\
$3$. Satellite Dishes \\
$4$. Even the McDonald Arch \\
Etc. Parabola's also form in nature, an example of this would be water coming out of water fountains,they form a parabola. Even when you throw a ball it forms a parabola (Projectile). \\
\subsection{Let's see how to do graph these functions?}
Equation $1:$ \\
$f(x)=x^2-7x+12$ \\
\includegraphics[width=0.7\textwidth]{hey.png}
Step $1$. We need to find the axis of symmetry \\
\textbf{Axis of Symmetry Formula}: $$x=\frac{-b}{2a}$$ \\
Step $2$. Plug in the values
a$=1$ and b$=-7$
$$x=\frac{-(-7)}{2(1)}$$
$$x=\frac{7}{2}$$
$$x=3.5$$
x is going to be $3.5$\\
step $3$. Plug in x into the original functions
$$f(x)=x^2-7x+12$$
$$f(3.5)=(3.5)^2-7(3.5)+12$$
$$f(3.5)= 12.3-24.5+12$$
$$f(3.5)=(-0.2)$$ \\
Step $4$. Identifying the vertex\\
Now we have both the x and y value, we can now find the vertex.
$$(x,y)=(3.5,(-0.2))$$
SO far, we know the
axis of symmetry which is $x=3.5$
vertex which is $(3.5,-0.2)$
We still need to find out the x-intercept. \\
\underline{x-intercept}
we can find this out when $y=0$
We can look at the graph and see that there are two x-intercepts, one is $3$ and the other is going to be $4$
This matches with our answer in which we used the quadratic equation to find. \\
If we wanted to find the $x$ value with a given $y$ value, we can do that as well. Say for example we wanted to find:
$$f(x) = 6$$
Using the graph for this equation, we can see that the $y$ value of 6 has two $x$ values: $1$ and $6$.\\
But our eyes may not be accurate enough, let's use math and see what the real value of $x$ is. We simply replace the $f(x)$ with $6$ in the original equation as they are equal to each other.
$$6 = x^2 - 7x + 12$$
Now it's simple algebra
$$0 = x^2 - 7x + 6$$
$$0 = (x - 6)(x - 1)$$
Therefore:
$$6 = f(6), f(1)$$
Equation $2$.
$$f(x)=4x^2+20x+25$$
\includegraphics[width=0.7\textwidth]{hey2.png}
We got this graph by first finding out the axis of symmetry
$$x=\frac{-b}{2a}$$
Step $2$. Plug in the values
since a= $4$ and b=$20$
$$x=\frac{-(20)}{2(4)}$$
$$x=\frac{-20}{8}$$
$$x=-2.5$$ \\
Step $3$. Plug in x
$$f(-2.5)=4(-2.5)^2+20(-2.5)+25$$
$$f(-2.5)=25+(-50)+25$$
$$f(-2.5)=0$$
So y=$0$.
\underline{x-intercept}
We already found out x-intercept since y=$0$, so therefore $-2.5$ is going to be the x-intercept. \\
What is $f(x) = 9?$
Same as above, we replace $f(x)$ with $9$
$$9 = 4x^2 + 20x + 25$$
$$0 = 4x^2 + 20x + 16$$
$$0 = 4x^2 + 4x + 16x + 16$$
$$0 = 4x(x + 1) + 16(x + 1)$$
$$0 = (4x + 16)(x + 1)$$
$$0 = 4(x + 4)(x + 1) = (x + 4)(x + 1)$$
Therefore:
$$9 = f(-4), f(-1)$$
Equation 3
$f(x)=-3x^2-5x+7$
This parabola will be facing downwards since $x^2$ has a - coefficient
\includegraphics[width=0.7\textwidth]{hey3.png}
We got this graph by first finding out the vertex.\\
$$x=\frac{-b}{2a}$$
A=$-3$ and B=$-5$
$$x=\frac{-(-3)}{2(-5)}$$
$$x=\frac{3}{-10}$$
$$x=-0.3$$\\
Now we use the axis of symmetry value to find out the f(x)
$$f(x)=-3x^2-5x+7$$
$$f(x)=-3(-0.3)^2-5(-0.3)+7$$
$$f(x)=-0.3-(-1.5)+7$$
$$f(x)=8.2$$
The vertex for this graph is going to be $(-0.3,8.2)$
Now we can graph this equation
The parabola is going going downwards since the first coefficient is a negative number and thus the parabola is going downwards.
The x-intercept is going to be $(-2.5)$ and $(0.9)$\\
Finally, let's see what $f(x) = 7$ is:
$$7 = -3x^2 - 5x + 7$$
$$0 = -3x^2 - 5x + 0$$
$$0 = -3x^2 - 5x$$
In this case, we can substitute $x$ with $0$ to get $0$.\\
Therefore
$$7 = f(0)$$
\subsubsection{Lets see how this graph works}
We can see how this works by plugging a number for f(x) and we can see the answer.
Equation $1$ \\
$1. f(x)=x^2-7x+12 \\$
So let f(x)=$3$, we look at the number $3$ on the y-Axis
\includegraphics[width=0.7\textwidth]{hey.png}\\
We can see that $5.3$ and $1.69$ came up from the x-axis. That is the answer
Equation $2$\\
$$2. f(x)= 4x^2+20x+25 \\$$
Now we can use in f(x)=$2$; we can look at the number $2$ on the y-axis.
\includegraphics[width=0.7\textwidth]{hey2.png}
We can see that the x value needed to make sure that f(x)=$2$ can be either $-1.79$ or $-3.2$.
How does this work?
There seems to be two answers and not just one?
Well it's because that a parabola is a special case and has $2$ roots not just one.
Equation $3$
$3. f(x)=-3x^2-5x+7$\\
\underline{Conclusion}\\
Quadratic functions can be seen everywhere in nature; from the arc of a fountain to the flight time of a ball thrown up into the air. Any object with parabolic properties will have a quadratic function related to it. A function where one input spits out two outputs, quadratic functions are truly unique and one of a kind.
\end{document}
```

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