Quadratic Functions — Parabolas, Key Features, Domain and Range

A quadratic function is a polynomial where the highest power of the input variable is \(2\).

A common form is

\[f(x)=ax^2+bx+c,\quad a\ne 0,\quad a,b,c\in\mathbb{R}.\]

The simplest quadratic is \(f(x)=x^2\). Its graph is a parabola.

Finding Domain and Range from a Sketch

• The domain is the set of allowed \(x\)-values (inputs).

• The range is the set of resulting \(y\)-values (outputs).

Worked Example — Range with a restricted domain

Let \(f(x)=7-2x-x^2\).

a. Sketch the graph for \(-5\le x\le 3\) and determine the range.

b. State the range if the domain were unrestricted.

Solution

First note \(f(x)=-(x^2+2x)+7\), so the parabola opens downward (there is a maximum).

The axis of symmetry is

\[x=-\frac{b}{2a}=-\frac{-2}{2(-1)}=-1,\]

so the vertex occurs at \(x=-1\). Then

\[f(-1)=7-2(-1)-(-1)^2=7+2-1=8,\]

giving a maximum value \(y=8\) at \((-1,8)\).

For the restricted domain, check the endpoints:

\[f(-5)=7-2(-5)-(-5)^2=7+10-25=-8,\]

\[f(3)=7-2(3)-3^2=7-6-9=-8.\]

So on \(-5\le x\le 3\), the outputs run from \(-8\) up to \(8\). Therefore:

\[-8\le y\le 8.\]

For the unrestricted domain (\(x\in\mathbb{R}\)), the parabola still has maximum \(8\) and falls without bound, so:

\[y\le 8.\]

Quadratics in Context — Choosing a Reasonable Domain and Range

In modelling, the domain and range are often limited by what is physically meaningful (time can’t be negative, heights usually aren’t negative, lengths must be positive, etc.).

Worked Example — Projectile height and maximum value

A projectile’s height above the ground (in metres) after \(t\) minutes is modelled by

\[h(t)=-0.846t^2+22.2t+478.\]

Sketch the graph in a sensible window and find the maximum height reached.

Solution

Since the input is time, it is reasonable to take

\[t\ge 0.\]

Also, height above ground is not negative while the projectile is in the air, so we only keep the part of the graph where \(h(t)\ge 0\).

The parabola opens downward (\(a=-0.846<0\)), so the maximum occurs at the vertex:

\[t_{\text{vertex}}=-\frac{b}{2a}=-\frac{22.2}{2(-0.846)}=\frac{22.2}{1.692}\approx 13.1.\]

Substitute \(t\approx 13.1\):

\[h(13.1)\approx -0.846(13.1)^2+22.2(13.1)+478\approx 624.\]

Therefore the maximum height is

\[h_{\max}\approx 624\text{ m} \quad \text{(at about } t\approx 13.1\text{ minutes).}\]

A suitable sketch would start at \((0,478)\), rise to near \((13.1,624)\), then fall back to the ground at about \(t\approx 40.3\) minutes.

Problems Involving Quadratics — Building a Model from Geometry

Many “maximum/minimum” geometry problems lead to a quadratic model. A typical workflow is:

(1) choose variables,

(2) write a constraint equation,

(3) express the quantity of interest as a function,

(4) interpret intercepts/vertex.

Worked Example — Maximizing area with a fixed perimeter

A rectangular mirror has perimeter \(260\text{ cm}\).

a. If the length is \(x\text{ cm}\), express the height in terms of \(x\).

b. Write the area \(A\text{ cm}^2\) in terms of \(x\).

c. Sketch/plot a graph of \(A\) against \(x\) with a suitable domain and range.

d. Find the \(x\)-intercepts of the graph and interpret them.

e. Find the equation of the axis of symmetry and explain what it means here.

Solution

a. Let the height be \(y\text{ cm}\). The perimeter condition is

\[2x+2y=260 \quad\Rightarrow\quad x+y=130 \quad\Rightarrow\quad y=130-x.\]

b. Area \(A=\text{length}\times\text{height}=xy\), so

\[A=x(130-x)=130x-x^2.\]

Hence

\(A(x)=130x-x^2\).

c. In context, both sides must be non-negative, so

\[0\le x\le 130.\]

The parabola opens downward, and its vertex occurs halfway between the zeros (see part d), at \(x=65\). Then

\[A(65)=130(65)-65^2=8450-4225=4225.\]

A sensible viewing window is \(0\le x\le 130\) and \(0\le A\le 4500\).

d. The \(x\)-intercepts satisfy \(A=0\):

\[x(130-x)=0 \Rightarrow x=0 \text{ or } x=130.\]

So the intercepts are \((0,0)\text{ and }(130,0)\).

These represent “degenerate” rectangles where one side is \(0\text{ cm}\) (no area), so they are the lower/upper bounds for possible \(x\)-values.

e. The axis of symmetry is halfway between \(0\) and \(130\):

\[x=65.\]

In this context, \(x=65\) is the length that produces the maximum area of the mirror.

Restricted Domains and Inverse Functions

Recall these key facts about inverses \(f^{-1}\):

• \(f^{-1}(f(x))=x\) and \(f(f^{-1}(x))=x\) (when the inverse exists).

• The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\).

• To find an inverse algebraically: write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject.

Worked Example — Restricting a quadratic to create an inverse

Consider \(f(x)=(x-2)^2\).

a. Explain why \(f^{-1}\) does not exist if the domain is \(x\in\mathbb{R}\).

b. The domain is restricted to \(x\ge k\). Find the smallest possible value of \(k\) so that \(f\) is invertible.

c. With this restriction, sketch \(f^{-1}\).

Solution

a. Over all real \(x\), the graph is a parabola symmetric about \(x=2\), so most \(y\)-values correspond to two different \(x\)-values.

For example,

\[f(0)=(0-2)^2=4,\]

\[f(4)=(4-2)^2=4,\]

so the function is many-to-one. Therefore \(f^{-1}\text{ does not exist as a function on }\mathbb{R}.\)

b. To make the function one-to-one, keep only one side of the parabola. The smallest \(k\) that works is the vertex \(x=2\).

Hence the restricted domain is

\[\{x\mid x\ge 2\} \quad \text{so } k=2.\]

c. Start with \(y=(x-2)^2\) and \(x\ge 2\). Swap \(x\) and \(y\):

\[x=(y-2)^2.\]

Taking square roots gives \(y-2=\pm\sqrt{x}\), but since we kept \(x\ge 2\) on \(f\), the inverse must output \(y\ge 2\), so we choose the positive branch:

\[f^{-1}(x)=2+\sqrt{x},\quad x\ge 0.\]

The graph of \(f^{-1}\) is the reflection of the right-hand branch of \(f\) in the line \(y=x\).

Helpful reflected point pairs include

\((2,0)\leftrightarrow(0,2)\),

\((3,1)\leftrightarrow(1,3)\),

\((4,4)\leftrightarrow(4,4)\),

\((5,9)\leftrightarrow(9,5)\).