Quadratic Functions and Conics

Section 2.2 Quadratic Functions and Conics

Art is not a mirror to reflect the world, but a hammer with which to shape it. — Vladimir Vladimirovich Mayakovsky, Russian poet, playwright, artist, and stage and film actor, 1893 –1930

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Problem: A hockey team plays in an arena that has a seating capacity of

$15,000$ spectators. With the ticket price set at

$\$50\text{,}$ average attendance at recent games has been

$10,500\text{.}$ A market survey indicates that for each dollar the ticket price is lowered, the average attendance increases by

$100\text{.}$

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Find a function that models the revenue in terms of ticket price.

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Solution

Let $x$ represent the ticket price in dollars and let $a(x)$ represent the average attendance at the price $x\text{.}$

Observe that, since for each dollar the ticket price is lowered, the average attendance increases (or decreases, if the price goes over $\$ 50$) by $100\text{,}$ the attendance with the ticket price of $x$ dollars is given by

\begin{equation*} a(x)=10500+100(50-x)=100(155-x). \end{equation*}

Note that the domain of the function $a(x)$ is the price range that makes sense in this particular situation. In other words, the price $x$ should satisfy these two conditions:

The price should not be lower than the price, call it $p\text{,}$ that guarantees the full arena, i.e. $15,000$ spectators in the audience.

To find the value of $p$ we solve the equation $a(p)=100(155-p)=15000\text{.}$ It follows that $p=5\text{.}$

The price should be lower than the price, call it $P\text{,}$ that would make the arena empty.

To find the value of $P$ we solve the equation $a(P)=100(155-P)=0\text{.}$ It follows that $P=155\text{.}$

This means that $x\in [5,155)\text{.}$

The revenue is given as the product of the ticket price and the number of the people in the attendance.

Hence the function

\begin{equation*} R(x)=x\cdot a(x)=x\cdot100(155-x)=100(155x-x^2) \end{equation*}

models the revenue $R(x)$ in terms of the ticket price $x\text{.}$

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Important observation: It is natural to ask the following question: Which value of

$x$ will give the maximum value of

$R(x)\text{?}$ Calculus will give you a simple tool to answer this type of questions.

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Quadratic Function: A quadratic function is a function

$f$ of the form

$\begin{equation*} f(x)=ax^2+bx+c \end{equation*}$

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where

$a\text{,}$

$b\text{,}$ and

$c$ are real numbers with

$a\not=0\text{.}$

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Example 2.2.1. Quadratic or not.

Which of the following functions is NOT a quadratic function:

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

Solution

The function $\displaystyle f(x)= \frac{1}{x^2+2x+1}$ is not quadratic because it is not in the form $y=ax^2+bx+c\text{.}$

Later in the course we will learn that $f$ is a so–called rational function.

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Warm Up:

Check that for any $a,b,c,x\in \mathbb{R}\text{,}$ $a\not=0\text{,}$

$\begin{equation*} ax^2+bx+c=a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}. \end{equation*}$
Solution

Note that

\begin{equation*} a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}=a\left( x^2+2\cdot x\cdot \frac{b}{2a}+\left( \frac{b}{2a}\right)^2\right)-\frac{b^2-4ac}{4a} \end{equation*}

\begin{equation*} =ax^2+bx+a\cdot \frac{b^2}{4a^2}-\frac{b^2-4ac}{4a}=ax^2+bx+\frac{b^2}{4a}-\frac{b^2-4ac}{4a} \end{equation*}

\begin{equation*} =ax^2+bx+\frac{b^2-(b^2-4ac)}{4a}=ax^2+bx+\frac{b^2-b^2+4ac}{4a} \end{equation*}

\begin{equation*} =ax^2+bx+\frac{4ac}{4a}=ax^2+bx+c. \end{equation*}

Important: You are expected to know the following identities

$(x+y)^2=x^2+2xy+y^2$ — the square of a sum

$(x-y)^2=x^2-2xy+y^2$ — the square of a difference
Complete the square, i.e. re–write the function $f(x)=2x^2-5x+2$ in the form

$\begin{equation*} f(x)=A(x+B)^2+C. \end{equation*}$
Solution

We use the previous example.

Note that if $f(x)=2x^2-5x+2$ then $a=2\text{,}$ $b=-5\text{,}$ and $c=2\text{.}$

This implies that

$A=a\text{,}$ $B=\frac{b}{2a}=\frac{-5}{4}=-\frac{5}{4}\text{,}$ and $C=-\frac{b^2-4ac}{4a}=-\frac{25-4\cdot 2\cdot 2}{4\cdot 2}=-\frac{9}{8}\text{.}$

Hence

\begin{equation*} f(x)=2x^2-5x+2=2\left( x-\frac{5}{4}\right)^2 -\frac{9}{8}. \end{equation*}
Draw a graph of the function $f(x)=2x^2-5x+2\text{.}$
Solution
From

\begin{equation*} f(x)=2x^2-5x+2=2\left( x-\frac{5}{4}\right)^2 -\frac{9}{8} \end{equation*}

we observe that we can obtained the graph of $f$ by using transformations of functions:
- Step 1. Start with the function $g(x)=x^2\text{.}$
- Step 2. Obtain the graph of $\displaystyle h(x)=g\left( x-\frac{5}{4}\right)=\left( x-\frac{5}{4}\right)^2$ by horizontally shifting the graph of $g$ by $\displaystyle \frac{5}{4}$ to the right.
- Step 3. Obtain the graph of $\displaystyle i(x)=2\cdot h(x)= 2\cdot \left( x-\frac{5}{4}\right)^2$ by vertically stretching the graph of $h$ by factor 2.
- Step 4. Obtain the graph of $\displaystyle f(x)=i(x) -\frac{9}{8}= 2\cdot \left( x-\frac{5}{4}\right)^2 -\frac{9}{8}$ by vertically shifting the graph of $i$ downwards by $\displaystyle \frac{9}{8}\text{.}$
See Figure 2.14.

Figure 2.14. The graph of $\displaystyle f(x)=2x^2-5x+2=2\left(x-\frac{5}{4}\right)^2-\frac{9}{8}\text{.}$

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All You Need to Know About Quadratic Functions: Let a quadratic function

$\begin{equation*} f(x)=ax^2+bx+c=a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}, a,b,c\in \mathbb{R}, a\not=0 \end{equation*}$

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be given.

What is the domain of the function $f\text{?}$
Solution
The domain of the function $f$ is the set $\mathbb{R}\text{,}$ the set of all real numbers.
Find the zeros of the function $f\text{,}$ i.e., solve the equation $f(x)=0\text{.}$

Solution

From

\begin{equation*} f(x)=0\Leftrightarrow a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}=0\Leftrightarrow a\left( x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a} \end{equation*}

we conclude that

\begin{equation*} f(x)=0\Leftrightarrow \left( x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}. \end{equation*}

Note that $\displaystyle \left( x+\frac{b}{2a}\right)^2\geq 0$ for all $x\in \mathbb{R}\text{.}$ Also since $4a^2\gt 0$ we have that

\begin{equation*} \frac{b^2-4ac}{4a^2}\geq 0 \Leftrightarrow b^2-4ac\geq 0. \end{equation*}

Summary:

$\begin{equation*} \begin{array}{l|l} b^2-4ac\gt 0\amp \text{Two solutions: }\displaystyle x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ and }x=\frac{-b-\sqrt{b^2-4ac}}{2a}\\ \hline b^2-4ac=0\amp \text{One solution: }\displaystyle x=-\frac{b}{2a}\\ \hline b^2-4ac\lt 0\amp \text{No solution}\\ \end{array} \end{equation*}$
Evaluate $\displaystyle f\left(-\frac{b}{2a}\right)\text{.}$
Solution

From

\begin{equation*} f(x)=a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a} \end{equation*}

it follows that

\begin{equation*} f\left(-\frac{b}{2a}\right)=a\left( -\frac{b}{2a}+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a} =-\frac{b^2-4ac}{4a}. \end{equation*}
Check the following two facts and discuss the intervals of increase and decrease in each case:
1. If $a\ge 0$ then, for all $\displaystyle x\not=-\frac{b}{2a}\text{,}$ $\displaystyle f(x)\gt -\frac{b^2-4ac}{4a}\text{.}$
  Solution
  
  We observe that if $\displaystyle x\not=-\frac{b}{2a}$ then $\displaystyle x+\frac{b}{2a}\not=0$ and
  
  \begin{equation*} \left( x+\frac{b}{2a}\right)^2>0. \end{equation*}
  
  This together with $a\gt 0$ implies that
  
  \begin{equation*} a\cdot \left( x+\frac{b}{2a}\right)^2>0. \end{equation*}
  
  Consequently
  
  \begin{equation*} f(x)=a\left( x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}>-\frac{b^2-4ac}{4a}. \end{equation*}
  
  In other words, if $\displaystyle x\not=-\frac{b}{2a}$ then
  
  \begin{equation*} f(x)>f\left(-\frac{b}{2a}\right) \end{equation*}
  
  or, for all $x\in \mathbb{R}\text{,}$
  
  \begin{equation*} f(x)\geq f\left(-\frac{b}{2a}\right). \end{equation*}
2. If $a\lt 0$ then, for all $\displaystyle x\not=-\frac{b}{2a}\text{,}$ $\displaystyle f(x)\lt -\frac{b^2-4ac}{4a}\text{.}$
  Solution
  Use the same reasoning as the above to prove the required inequality.

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Putting everything together: See Figure 2.15 and Figure 2.16.

Let $a\gt 0\text{.}$ Then

Figure If $a\gt 0$ then the graph of $f$ is a “smiley parabola”.
Let $a\lt 0\text{.}$ Then

Figure If $a\lt 0$ then the graph of $f$ is a “sad parabola”.

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Fast Motion: Draw a graph of the function

$f(x)=-x^2+x+2\text{.}$

$\begin{equation*} \begin{array}{c|c|c|c} \text{The }x \text{-intercepts:}\amp \text{The sign of }a\amp \text{The vertex }\displaystyle \left(-\frac{b}{2a},-\frac{b^2-4ac}{4a}\right)\amp \text{The }y \text{-intercept}\\ \text{solve }f(x)=0\amp \amp \amp\\ \hline -x^2+x+2=0\amp \amp \amp f(0)=2\\ \displaystyle x=\frac{-1\pm \sqrt{1+8}}{-2}\amp a=-2\lt 0\amp \displaystyle \left(-\frac{1}{-2},-\frac{1-4(-1)(2)}{-4}\right)\amp \\ \displaystyle x=\frac{-1\pm 3}{-2}\amp \text{Sad parabola}\amp \displaystyle \left(\frac{1}{2},\frac{9}{4}\right)\amp\\ x=-1\text{ or }x=2\amp \amp \amp \\ \end{array} \end{equation*}$

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See Figure 2.17.

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Figure 2.17. Graph of

$f(x)=-x^2+x+2\text{.}$

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Meet … Conics: Watch

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and

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Circle:

A circle is the set of all points in a plane that are at a given distance from a given point, the centre.The distance between any of the points and the centre is called the radius of the circle.

See Figure 2.18.

Figure 2.18. A circle.
Evaluate the distance between the two points, $A=(x_1,y_1)$ and $B=(x_2,y_2)\text{,}$ i.e., find the length of the line segment $\overline{AB}\text{.}$

See Figure 2.19.

Figure 2.19. We use the Pythagorean theorem – the most famous theorem of all time.
By the Pythagorean Theorem the distance between $A=(x_1,y_1)$ and $B=(x_2,y_2)$ is given by:

$\begin{equation*} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \end{equation*}$

YOU MUST KNOW THE DISTANCE FORMULA!
The equation of the circle with the centre at the point $(a,b)$ and the radius $r$ is given by

$\begin{equation*} (x-a)^2+(y-b)^2=r^2. \end{equation*}$

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Example 2.2.2. Circles.

Draw the following circles:
- The unit circle: $x^2+y^2=1\text{,}$
- $(x-2)^2+y^2=2\text{,}$
- $(x-1)^2+(y-1)^2=2\text{.}$
Write the equation of the circle with the centre at the point $(-1,1)$ and radius $r=\pi\text{.}$
Find the centre and the radius of the circle $x^2+2x+y^2-6y=0\text{.}$

Solution

The unit circle is a circle centred at the origin and a radius equal to 1.

The circle given by $(x-2)^2+y^2=2$ is centred at $(2,0)$ with the radius $r=\sqrt{2}\text{.}$

The circle given by $(x-1)^2+(y-1)^2=2$ is centred at $(1,1)$ with the radius $r=\sqrt{2}\text{.}$

See Figure 2.20.

Figure 2.20. Three circles.
An equation of the circle with the centre at the point $(-1,1)$ and radius $r=\pi$ is given by $(x+1)^2+(x-1)^2=\pi^2\text{.}$
Observe that

\begin{equation*} x^2+2x+y^2-6y=0\Leftrightarrow (x^2+2x+1)-1+(y^2-6y+9)-9=0\Leftrightarrow (x+1)^2+(y^2-3)^2=10. \end{equation*}

Hence the centre of the circle is at the point $(-1,3)$ and the radius is $r=\sqrt{10}\text{.}$

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Elipse:

An ellipse is the set of all points in a plane with the property that the sum of the distances to two given points equals to the given number.

See Figure 2.21.

Figure 2.21. An ellipse
Many ways to draw an ellipse:
The standard form of the equation of an ellipse centred at the origin is

$\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{equation*}$

See Figure 2.22.

Figure The ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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Hyperbola:

A hyperbola is the set of all points in a plane with the property that the absolute value of the difference of the distances to two given points equals to the given number.

See Figure 2.23.

Figure 2.23. A hyperbola
Many ways to draw an hyperbola:
The standard form of the equation of an ellipse centred at the origin is

$\begin{equation*} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{equation*}$

See Figure 2.24.

Figure The ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
The standard form of the equation of a hyperbola centred at the origin is

$\begin{equation*} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \end{equation*}$

See Figure 2.24.

Figure The hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$