3.2 Linear and quadratic functions

A line models a quantity that changes by the same amount whenever its input increases by one. A quadratic function models a change whose rate itself changes; its graph turns at a highest or lowest point. Recognising these two shapes helps you interpret costs, distances, areas, and scores without treating every graph as a new kind of problem.

This is foundational to intermediate algebra. Module 2.2 explains how to solve a line's equation; Module 2.3 explains how to find a quadratic's roots. Here those answers become points on graphs. Review linear functions before moving to quadratic functions in OpenStax.

A basic idea: constant change or a turning point

The line $L(x)=2x+1$ rises by $2$ whenever $x$ rises by $1$. Its graph crosses the vertical axis at $1$. The quadratic $Q(x)=-(x-2)^2+5$ rises, reaches $5$ at $x=2$, then falls. Inputs equally far from $2$ give the same output: $Q(1)=Q(3)=4$.

A quick tour of formulas

For a line,

$$L(x)=mx+b,\qquad m=\frac{L(x_2)-L(x_1)}{x_2-x_1}\quad(x_2\ne x_1).$$

The slope $m$ is output change per unit of input; $b=L(0)$ is the vertical intercept. A positive slope rises, a negative slope falls, and a zero slope is flat. Two points with different inputs determine one line.

A quadratic can be written in several useful forms:

$$Q(x)=ax^2+bx+c =a(x-h)^2+k,\qquad a\ne0.$$

In vertex form, $(h,k)$ is the turning point. In standard form, $h=-b/(2a)$ and $k=Q(h)$. When $a>0$ the vertex is a minimum; when $a<0$ it is a maximum. The completing-the-square method explains where vertex form and $h=-b/(2a)$ come from. If a quadratic can be factored as $a(x-r_1)(x-r_2)$ with real $r_1,r_2$, its zeros are $r_1$ and $r_2$. A parabola need not meet the horizontal axis: when $b^2-4ac<0$ there are no real zeros.

Check the domain supplied by a word problem. A formula may accept negative inputs even when a distance, time, or count cannot.

Application 1: a fare with a fixed charge

A ride costs $30$ units to start and $8$ units per kilometre. What is the cost of $7$ kilometres, and what does each part of the graph mean?

1. Write the line. For distance $d\ge0$, $C(d)=30+8d$.
2. Evaluate. $C(7)=30+8(7)=86$ units.
3. Read the graph. The vertical intercept is $30$, the cost at zero distance. The slope is $8$ units per kilometre. Any 3 km increase, such as from $d=2$ to $d=5$, adds $3\cdot8=24$ units, whatever the starting distance.

The straight graph is evidence that the rate stays constant.

Application 2: a rectangle with fixed side sum

Two adjacent sides of a rectangle must add to $12$ units. If one side is $x$, what is the largest possible area?

1. Write the valid inputs and area. The other side is $12-x$, so $A(x)=x(12-x)=-x^2+12x$, with $0<x<12$.
2. Find the vertex. Here $a=-1$ and $b=12$, so $h=-b/(2a)=6$. The area is $A(6)=6(6)=36$ square units.
3. Check the meaning. The parabola opens downward, and $x=6$ is within the domain. The two sides are both $6$. Values $x=4$ and $x=8$ each give $32$ square units, illustrating symmetry about $x=6$.

The quadratic graph shows both the maximum and why moving either side away from $6$ reduces the area.