Lecture 2 · Barnett 1-3 · about 45 min

Linear and Quadratic Functions

Lines from a slope or two points, break-even with linear cost and revenue, and the parabola: vertex, axis, intercepts, vertex form, and the quadratic revenue and profit models that come from a linear demand curve.

By the end you can

Write a linear function from a slope and a point, or from two points, and interpret the slope economically
Find and draw the break-even point of linear cost and revenue
Sketch a parabola from its vertex, axis and intercepts, and write it in vertex form
Build revenue from a demand equation and find maximum revenue, maximum profit and break-even points

Key idea · Exam recipe (memorise this)

Line through two points: $m = \frac{y_2 - y_1}{x_2 - x_1}$ , then $y - y_1 = m(x - x_1)$ , solve for $y$ . Parallel: same $m$ . Perpendicular: $m_1 m_2 = -1$ . Horizontal $y = b$ (slope 0), vertical $x = a$ (no slope).
Linear break-even: $R = px$ , $C = F + vx$ , set $R = C$ : $x = \frac{F}{p - v}$ . Profit $P = (p - v)x - F$ .
Parabola $f(x) = ax^2 + bx + c$ : sign of $a$ (up/min or down/max) → vertex $h = -\frac{b}{2a}$ , $k = f(h)$ → axis $x = h$ → y-intercept $c$ → x-intercepts (factor or $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ) → range $[k, \infty)$ or $(-\infty, k]$ .
Vertex form $a(x - h)^2 + k$ : take $a$ out of the $x$ terms, complete the square inside, multiply back.
Demand to revenue: $R(x) = x \cdot p(x)$ . Max revenue at the vertex of $R$ .
Max profit: $P(x) = R(x) - C(x)$ , vertex of $P$ . Break-even: $P(x) = 0$ (two roots, profit between them).

How the exam asks it: three core exam templates

In midterm exams, linear and quadratic questions focus on three canonical structures: building linear cost and break-even models, sketching and converting parabolas, and optimising quadratic revenue/profit from linear demand. Master these three worked solutions, then follow the cheat-sheet and 4-step method below.

Exam template 1: Linear cost specification and break-even analysis

Question: A manufacturer produces handcrafted bags. Total cost is 800 TL for 50 bags and 1,100 TL for 100 bags. Each bag sells for $p = 50$ TL. Separately, another product line has cost $C(x) = 30x + 4000$ and sells at 50 TL each.
a) Assuming cost is linear, find the slope $m$ and determine the cost function $C(x)$ .
b) Identify the fixed cost and variable cost per unit from $C(x)$ .
c) For the second product line ( $C(x) = 30x + 4000$ , $p = 50$ TL), find the break-even quantity and the break-even dollar revenue.
d) Calculate the profit earned from the second product line when 300 units are produced and sold.

Exam template 2: Parabola sketching, vertex coordinates, and vertex form

Question: Consider the quadratic function $f(x) = x^2 - 4x - 5$ .
a) Find the vertex $(h, k)$ and state the equation of the axis of symmetry.
b) Find the coordinates of the $y$ -intercept and all $x$ -intercepts.
c) State whether $f(x)$ has a maximum or minimum, and write its range in interval notation.
d) Write $g(x) = -2x^2 + 12x - 10$ in vertex form $a(x - h)^2 + k$ by completing the square, and identify its vertex.

Exam template 3: Quadratic revenue, profit maximisation, and break-even points

Question: A firm faces price-demand equation $p = 100 - 2x$ and total cost $C(x) = 20x + 600$ , where $x$ is output.
a) Find the total revenue function $R(x)$ and state its economic domain.
b) Formulate the total profit function $P(x)$ .
c) Find the break-even quantities where the firm earns zero profit.
d) Find the production quantity $x$ that maximises profit, the maximum profit amount, and the corresponding price charged per unit.

Cheat-sheet & the 4-step method

· Cheat-sheet: linear equations, break-even, and quadratic parabolas

Model / Question	Formula or Condition	When to use it	How to write the exam answer
Slope between 2 points	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$	Two known cost or demand data points	Compute $m$ ; interpret as marginal cost or price change per unit
Point-slope equation	$y - y_1 = m(x - x_1)$	Building linear equation from point and slope	Substitute $(x_1, y_1)$ and $m$ , rearrange to $y = mx + b$
Linear break-even	$R(x) = C(x) \implies x_{BE} = \dfrac{F}{p - v}$	Finding zero-profit point for linear revenue and cost	State formula, compute $x_{BE}$ , substitute into $R(x)$ for dollar volume
Parabola vertex	$h = -\dfrac{b}{2a}, \quad k = f(h)$	Finding maximum/minimum of $ax^2 + bx + c$	$h$ is optimal quantity, $k$ is maximum profit/revenue or min cost
Parabola orientation	$a \gt 0$ opens up (min); $a \lt 0$ opens down (max)	Determining whether vertex is peak or valley	State sign of $a$ , conclusion (minimum or maximum)
Vertex form	$f(x) = a(x - h)^2 + k$	Completing square to reveal vertex directly	Complete square: factor $a$ , add/subtract $(B/2)^2$ , distribute
Quadratic formula	$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$	Finding break-even or $x$ -intercepts when not factorable	Compute discriminant $\Delta = b^2 - 4ac$ , solve both roots
Revenue from demand	$R(x) = x \cdot p(x)$	Linear demand $p = a - bx$ converted to revenue	Multiply by $x$ : $R(x) = ax - bx^2$ ; domain $0 \le x \le a/b$

Key idea · The 4-step method for any exam question on linear and quadratic models

Step 1: Identify the functional form and parameters.
- If linear: identify slope $m$ , intercept $b$ , fixed cost $F$ , variable cost $v$ , or unit price $p$ .
- If quadratic: identify coefficients $a, b, c$ and determine whether the parabola opens up ( $a \gt 0$ ) or down ( $a \lt 0$ ).
Step 2: Set up the governing formula.
- For break-even: set $P(x) = 0$ or $R(x) = C(x)$ .
- For maximum profit or revenue: set $h = -b/(2a)$ .
- For roots / intercepts: set $f(x) = 0$ .
Step 3: Perform arithmetic one operation at a time.
- Calculate discriminant $b^2 - 4ac$ before square-rooting.
- For vertex values, evaluate $k = f(h)$ carefully respecting negative signs (e.g. $-2(20)^2 = -800$ ).
Step 4: State the final bold exam sentence with units and verify.
- Distinctly label quantity ( $x$ , units) versus monetary value (TL, dollars).
- Check symmetry: $x$ -intercepts must be equidistant from the axis $x = h$ .

Linear functions

Slope-intercept formFormula sheet →

y = mx + b

$m$: slope, the change in y per unit change in x
$b$: y-intercept, the value of y when x = 0

Use it when you know the rate of change and the starting value. In a cost function, m is the variable cost per unit and b is the fixed cost.

Point-slope formFormula sheet →

y - y_1 = m(x - x_1), \qquad m = \frac{y_2 - y_1}{x_2 - x_1}

$(x_1, y_1), (x_2, y_2)$: two points on the line
$m$: slope

Use it when you are given two points or one point and a slope.

Break-even with two lines

Linear break-evenKey formulaFormula sheet →

R(x) = C(x) \;\Rightarrow\; px = F + vx \;\Rightarrow\; x_{BE} = \frac{F}{p - v}

$p$: price per unit
$v$: variable cost per unit
$F$: fixed cost
$x_{BE}$: break-even quantity

Use it when revenue and cost are both linear in x.

Quadratic functions: the parabola

Standard formFormula sheet →

f(x) = ax^2 + bx + c

$a$: shape: a > 0 opens upward (minimum), a < 0 opens downward (maximum)
$c$: y-intercept, f(0)

Use it when you are given a quadratic in expanded form and need its graph.

Vertex and axisKey formulaFormula sheet →

h = -\frac{b}{2a}, \qquad k = f(h), \qquad \text{axis: } x = h

$(h, k)$: the vertex
$k$: the minimum (a > 0) or maximum (a < 0) value

Use it when you need the optimum of a quadratic, or the centre of its graph, without derivatives.

The discriminant $b^2 - 4ac$ counts the x-intercepts: positive → two, zero → one (vertex on the axis), negative → none.

Quadratic formulaFormula sheet →

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

$a, b, c$: the coefficients of ax^2 + bx + c
$b^2 - 4ac$: the discriminant

Use it when a quadratic does not factor easily and you need its roots (break-even points).

Revenue, cost and profit with a parabola

A linear demand $p = a - bx$ makes revenue $R = x \cdot p$ quadratic. With a linear cost, profit is quadratic too.

Raise the net price $b$ and the vertex moves up and right. Raise the fixed cost $c$ and the whole curve shifts down: the break-even points move toward each other, and if $c$ is too large there are none.

Exam practice

Summary and review

Line: $m = \frac{\Delta y}{\Delta x}$ , $y - y_1 = m(x - x_1)$ ; parallel same slope, perpendicular $m_1 m_2 = -1$ .
Linear break-even: $x = F/(p - v)$ ; draw $C$ from the fixed cost and $R$ from the origin.
Parabola: vertex $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$ , axis, intercepts, range; vertex form by completing the square.
$R(x) = x \cdot p(x)$ ; $P = R - C$ ; max profit at the vertex of $P$ , break-even at its roots.
Discriminant: two, one, or no x-intercepts.