Quadratic Functions

class: center, middle, inverse, title-slide

.title[
# Quadratic Functions
]
.subtitle[
## Session 05
]
.author[
### Alejandro Ucan
]
.date[
### 2023-10-01
]

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# Goals

* Describe the general form a the Quadratic function model. 
* Describe intercepts and vertices of the quadratic function. 
* Apply the quadratic function to model business problems.

---
# General Form of the Quadratic Function

> __Definition:__ The quadratic function is a polynomial function of degree 2. The general form of the quadratic function is given by: `$$f(x)=ax^2+bx+c$$` where `$a$`, `$b$` and `$c$` are real numbers and `$a\neq 0$`. The domain of the quadratic function is the set of all real numbers, but the range differs from one function to another.

---
## Its graph

> __Definition:__ The graph associated to a quadratic function is called a parabola. The parabola is symmetric with respect to the vertical line that passes through the vertex. The vertex is the point where the parabola changes direction. The parabola opens upward if `$a>0$` and downward if `$a<0$`.

---
## Its graph

<div class="figure" style="text-align: center">
<img src="index_files/figure-html/unnamed-chunk-1-1.png" alt="Parabola" width="100%" />
Parabola
</div>

---
## The intercepts with the x-axis.

> __Definition:__ The intercepts with the x-axis are the points where the parabola intersects the x-axis. The x-intercepts are the solutions of the equation `$f(x)=0$`. The x-intercepts are also called the zeros of the function.

--
In order to find our roots, we need to solve the equation `$f(x)=0$`. This equation is called the __quadratic equation__ and it has the form: `$$ax^2+bx+c=0,$$` we can use the quadratic general formula to solve that equation: `$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$` and dependending on the number `$\Delta=b^2-4ac$` we can have three different cases:
* `$\Delta>0$` two real roots.
* `$\Delta=0$` one real root.
* `$\Delta<0$` no real roots.

---
## The intercepts with the x-axis.

> __Definition:__ the number `$\Delta=b^2-4ac$` is called the __discriminant__ of the quadratic equation.

In order to interpretate the discriminant, we need to think that:
* If `$\Delta>0$` then the parabola intersects the x-axis in two different points.
* If `$\Delta=0$` then the parabola intersects the x-axis in one point.
* If `$\Delta<0$` then the parabola does not intersect the x-axis.

---
#### Examples:

> Determine the nature of the roots for each quadratic function: 
 * `$f(x)=x^2-4x+3$` 
 * `$f(x)=x^2-4x+4$` 
 * `$f(x)=x^2+2x+1$` 
 * `$f(x)=x^2+1$`

---
## The vertex of the parabola.

> __Definition:__ The vertex of the parabola is the point where the parabola changes direction. The vertex is the point `$(h,k)$` where `$h$` and `$k$` are the coordinates of the vertex. Usually, this point is where the parabola attains its minimum or maximum value.

--
To easily find my vertex, we need to write our parabola equation in the form: `$$f(x)=p(x-h)^2+k$$` where `$h$` and `$k$` are the coordinates of the vertex.

---
### The Square Completing process

Assume that we have the following quadratic function: `$$f(x)=ax^2+bx+c$$` and we want to write it vertex form. We can use the following process: 
1. Take the two first terms and factor out the coefficient of the squared term. `$$f(x)=a(x^2+\frac{b}{a}x)+c$$` 
1. Add and subtract the square of half the coefficient of the linear term. `$$f(x)=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c$$` 
1. Factor the first three terms and simplify. `$$f(x)=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$$` 
1. Simplify the last two terms. `$$f(x)=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$$`

---
#### Example 1:

> Write the quadratic function `$f(x)=x^2-4x+3$` in vertex form.
--

##### Solution:

`$$f(x)=x^2-4x+3=x^2-4x+4-4+3=(x-2)^2-1$$`

Note that here, `$a=1$`, `$b=-4$` and `$c=3,$` and `$\Delta=b^2-4ac=16-12=4>0$` and `$$h=\frac{-b}{2a}=\frac{4}{2}=2$$` and `$$k=\frac{-b^2+4ac}{4a}=\frac{16-12}{4}=1$$`

---
#### Example 2:

> Write the quadratic function `$f(x)=x^2+1$` in vertex form.
--

##### Solution:

`$$f(x)=x^2+1=x^2+1+0=(x+0)^2+1$$`

Note that here, `$a=1$`, `$b=0$` and `$c=1,$` and `$\Delta=b^2-4ac=0-4=4<0$` and `$$h=\frac{-b}{2a}=\frac{0}{2}=0$$` and `$$k=\frac{-b^2+4ac}{4a}=\frac{0+4}{4}=1$$`

---
# Applications

#### Example 3:

> The financial department in the company that produces a digital camere arrived at the following price-demand function and the following revenue function: `$$p(x)=94.8-5x$$` `$$R(x)=xp(x)=x(94.8.5x)$$` where `$p(x)$` measures the wholesale price per camera at which `$x$` million cameras can be sold, and `$R(x)$` is the corresponding revenue (in million of dollars). We are interested in the case where `$1\leq x\leq 15.$` 
 * Find the price that maximizes the revenue. 
 * Find the maximum revenue.

---
#### Example 4:

> The marketing research department for a company that manufactures and sells notebook computers stablished the following price-demand and revenue functions: `$$p(x)=2000-60x\quad R(x)=xp(x)=x(2000-60x)$$` where `$p(x)$` measures the wholesale price per computer at which `$x$` thousand computers can be sold, and `$R(x)$` is the corresponding revenue (in thousand of dollars). We are interested in the case where `$0\leq x\leq 25.$` 
 * Find the price that maximizes the revenue. 
 * Find the maximum revenue.