# MU123 Unit 10 Notes

# Introduction

Equation of the form:
$$
y = ax^2 +bx + c
$$
where $a$, $b$ and $c$ are constants with $a \ne 0$, are examined in this unit.
The *RHS* is a quadratic expression and graphs of these equations have a curved shape called *parabola*.

# 1 Introducing parabolas

## 1.1 Parabolas everywhere

### The free-fall equation

The relationship between the distance $d$ fallen by an object and the time $t$ that it has been falling is given by $$ d = \frac{1}{2}gt^2 $$ where the constant $g$ is the acceleration due to gravity, which is about $9.8 \text{m/s}^2$. Or more compactly: $$ d = 4.9t^2 $$

**Parabola** - a part of a parabola is formed by curved shape in the free-fall equation.
**Parabolic curve** - a curve whose shape is all or part of a parabola.

#### Definitions of a parabola

- A shape obtained when a plane "parallel" to the side of a cone intersects the cone.
- A shape of the graph of any equation of the form: $$ y = ax^2 +bx + c $$ where $a$, $b$ and $c$ are constants with $a \ne 0$. The free-fall equation has this form with $b = c = 0$.

The free-fall equation is only a **model** for the motion of a falling object. It's accurate when there are no other forces than gravity acting on the object.
In real life there actually are other forces, notably *air resistance* (*drag*) which slows down the falling object. After great enough distance the air resistance will cause the speed of falling object to stop increasing - to become constant. This is known as the *terminal velocity*.

**Quadratic model**
Any model based on an equation of form $y = ax^2 +bx + c$.

## 1.2 Projectiles

An object that is propelled through space by a force that ceases after launch.
**Trajectory** is the path a projectile follows.
**Ballistics** is the science of projectiles.

## 1.3 Stopping distances

# 2 Graphs of quadratic functions

Every parabola has an **axis of symmetry** cutting the parabola at exactly one point - the **vertex**.
A rule that takes input values and produces output values is called a **function**.

A function of the form $y = ax^2 + bx + c$ is called a **quadratic function**.

## 2.1 Graphs of equations of the form $y = ax^2$

The **magnitude** of $a$ affects how narrow the graph is. The larther the magnitude of $a$, the narrower the parabola becomes.

**u-shaped**parabola is same way up as the graph of $y = x^2$.**n-shaped**parabola is the another way up, e.g. $y = -x^2$.

The vertex is $(0, 0)$.

## 2.2 Graphs of equations of the form $y = ax^2 + bx + c$

The graph has the same shape as the graph of $y = ax^2$, but shifted and crosses the $y$-axis at $(0, c)$.

## 2.3 The intercepts of a parabola

The graph of $y = ax^2 + box + c$ can have two, one or zero $x$-intercepts. There is always exactly one $y$-intercept.

## 2.4 Sketch graphs of quadratic functions

- Find whether the parabola is u-shaped or n-shaped.
- Find its intercepts, axis of symmetry and vertex.
- Plot the features found, and hence sketch the parabola.
- Lable the parabola with its equation, and make sure that the values of the intercepts and the coordinates of the vertex are indicated.

A formula for the axis of symmetry of a parabola with equation $y = ax^2 + bx + c$ is: $$ x = - \frac{b}{2a} $$

# 3 Solving quadratic equations

## 3.2 The quadratic formula

The solutions of the quadratic equation $$ ax^2 + bx + c = 0 $$ are given by $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

### Simplifying a quadratic equation

- If the coeficient of $x^2$ is negative, then multiply the equation through by $-1$ to make this coefficient positive.
- If the coefficients have a common factor, then divide the equation through by this factor.
- If any of the coefficients are fractions, then multiply the equation through by a suitable number to clear them.

## 3.3 The number of solutions of a quadratic equation

The value $b^2 - 4ac$ is called the **discriminant** of the quadratic expression $ax^2 + bx + c$. Such an equation has:

- two solutions if $b^2 - 4ac > 0$ (the discriminant is positive).
- one solution if $b^2 - 4ac = 0$ (the discriminant is zero).
- no solutions if $b^2 - 4ac < 0$ (the discriminant is negative).

## 3.4 Vertically-launched projectiles

If an object is launched upwards from an initial height $h_0$ with an initial speed $v_0$, then after time $t$ its height $h$ is given by $$ h = -\frac{1}{2}gt^2 + v_0t + h_0 $$ where $g$ is the acceleration due to gravity, which is about $9.8\text{m}/\text{s}^2$.

# 4 Completing the square

Is a way to rearrange a quadratic expression.

## 4.1 Shifting parabolas

If a parabola with equation $y = ax^2$ is shifted right by $h$ units and up by $k$ units, then each point $(x, y)$ on the second parabola is a shift of the point $(x - h, y - k)$ on the first parabola. Hence the formula for second parabola is: $$ y - k = a(x - h)^2 $$ Or equivalently: $$ y = a(x - h)^2 + k $$ Which can be further multiplied out to give the usual form $y = ax^2 + bx + c$. Importantly the process works vice-versa too. Any equation of the form $y = ax^2 + bx + c$ where $a$, $b$ and $c$ are constants with $a \ne 0$, can be rearranged into the form $y = a(x - h)^2 + k$.

Every expression of the form $ax^2 + bx + c$, where $a \ne 0$, can be rearranged into the form
$$
a (x + \text{a number})^2 + \text{a number}
$$
where each of the two numbers in this expression can be positive, negative or zero.
This is called the **completed-square form**.

The parabola with equation $$ y = a(x - h)^2 + k $$ has vertex $(h, k)$.

## 4.2 Completing the square in quadratics of the form $x^2 + bx + c$

Where the coefficient of $x^2$ is 1 - the completed-square form looks like: $$ ( x + \text{a number} ) ^2 + \text{a number} $$

### Completing the square in quadratics of the form $x^2 + bx$

In general: $$ (x + p)^2 = x^2 + 2px + p^2 $$ The RHS expression is of the form $x^2 + bx$ plus an extra number. If I subtract this extra number from both sides and swap the sides, then I obtain: $$ x^2 + 2px = (x + p)^2 - p^2 $$ The RHS expression is the completed-square form of the LHS expression with following properties:

- Constant term in the RHS bracket is half of the coefficient of the term in $x$ on LHS.
- The number that is subtracted on the RHS is the square of the constant term in the brackets.

### Completing the square in quadratics of the form $x^2 + bx + c$

- Complete the square for the $x^2$ and $x$ terms as before.
- Collect the constant terms.

E.g.: $$ \begin{aligned} x^2 + 8x + 10 = \\[0.6em] (x + 4)^2 - 16 + 10 = \\[0.6em] (x + 4)^2 - 6 \end{aligned} $$

Hence the general strategy:

- Rewrite the expression with the $x^2 + bx$ part changed to $(x + p)^2 - p^2$, where the number $p$ is half of $b$.
- Collect the constant terms.

## 4.3 Solving quadratic equations by completing the square

- Equations, where the coefficient of $x^2$ is greated than 1, should be divided through by given coefficient.
- Complete the square.
- Rearrange the equation so the square term and the constant are on different sides.
- Take the square root of both sides.
- Rearrange the equation to obtain $x$ by itself on one side.

### The derivation of the quadratic formula

The general quadratic equation $ax^2 + bx + c = 0$ can be rearranged to obtain $x$ by itself on the LHS by following the steps:

Divide through by the coefficient of $x^2$: $$ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 $$ Complete the square: $$ (x + \frac{b}{2a})^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0 \\[1em] $$ Get the constant terms on the right, combine them into a single fraction: $$ \begin{aligned} (x + \frac{b}{2a})^2 &= \frac{b^2}{4a^2} - \frac{c}{a} \\[1em] &= \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \\[1em] &= \frac{b^2 - 4ac}{4a^2} \end{aligned} $$ Take the square root of both sides: $$ \begin{aligned} x + \frac{b}{2a} &= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \\[1em] &= \pm \frac{\sqrt{b^2 - 4ac}}{2a} \end{aligned} $$ Get $x$ by itself on the LHS: $$ \begin{aligned} x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \\[1em] &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{aligned} $$ This is the quadratic formula!

## 4.4 Completing the square in quadratics of the form $ax^2 + bx + c$

The common factor of $x^2$ and $x$ can be taken out of present, e.g.:

$$ 2x^2 + 8x -7 = 2(x^2 +4x) - 7 $$

Then the square in the qudratic *inside the brackets* can be completed as usual.

When there is no common factor, the coefficient of $x^2$ can still be taken out, the process will create fractions though:

$$ 2x^2 + 5x + 1 = 2(x^2 + \frac{5}{2}x) + 1 $$

### Strategy to complete the square in a qudratic of the form $ax + bx + c$

- Rewrite the expression with the coefficient $a$ of $x^2$ taken out of the $ax^2 + bx$ part as a factor. This generates a pair of brackets.
- Complete the square in the simple quadratic inside the brackets, remembering to keep it enclosed within its brackets. This generates a second pair of brackets, inside the first pair.
- Multiply out the
*outer*brackets. - Collect the constant terms.

# 5 Maximisation problems

Is the problem of finding the maximum value of a quantity and the circumstances under which it is obtained.

This section focuses on problems that can be modelled with a quadratic function, hence the solution is to find a vertex.

### Methods to find the vertex of a parabola from its equation

- Use the formula $x = -b / (2a)$ to find the $x$-coordinate, then substitute into the equation of the parabola to find the $y$-coordinate
- Find the $x$-intercepts (if there are any); then the value halfway between them is the $x$-coordinate of the vertex. Find the $y$-coordinate by substituting into the equation of the parabola.
- Complete the square: the parabola with equation $y = a(x - h)^2 + k$ has vertex $(h, k)$.
- Plot the parabola and read off the approximate coordinates of the vertex.

### Strategy to solve a maximisation problem

- Identify the quantity to be maximised and the quantity that it depends on, and denote each quantity by a variable.
- Find a formula for the variable to be maximised in terms of the variable that it depends on.
- If this gives a quadratic function, then find the vertex of its graph.