Part I · Mathematical Language, Functions, and Algebraic Models

Chapter 3 — Lines, Linear Models, and Quadratics

Why This Matters

Linear relationships describe proportional change and local approximations. Quadratic relationships describe curvature, extrema, and roots. Together they form the algebraic vocabulary needed for circuit equations, optimisation, and later calculus.

3.1 Straight lines and slope

A non-vertical straight line has equation

\[y=mx+b.\]

The constant $$m$$ is the slope: change in $$y$$ per unit change in $$x$$ . The constant $$b$$ is the $$y$$ -intercept, because $$x=0$$ gives $$y=b$$ .

For two distinct points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ on a non-vertical line,

m=\frac{y_2-y_1}{x_2-x_1},\qquad x_2\ne x_1.

Why does this work? The line equation gives $$y_1=mx_1+b$$ and $$y_2=mx_2+b$$ . Subtracting eliminates $$b$$ :

\[y_2-y_1=m(x_2-x_1).\]

Dividing by the nonzero horizontal change gives the slope formula. A vertical line has $$x_2-x_1=0$$ , so its slope is undefined and its equation is $$x=c$$ .

Slope is a ratio with direction. Moving right by a chosen run and measuring the signed rise gives the same quotient anywhere on a straight line. A negative slope does not mean a negative length; it records that the output decreases as the input increases.

m=\frac{\Delta y}{\Delta x}

Positive, zero, undefined, and negative slopes as oriented rise-over-run relationships.

3.2 Point-slope and two-point forms

Starting with $$y=mx+b$$ and a point $$(x_1,y_1)$$ on the line,

\[y-y_1=m(x-x_1).\]

This point-slope form is often the quickest way to write a line. Substituting the two-point slope gives

\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1},

provided the line is not vertical and $x\ne x_1$ during the displayed division. The undivided point-slope form remains safer in symbolic work.

Example — Constructing a calibration line

A sensor gives $0.4\ \mathrm V$ at $20^\circ\mathrm C$ and $2.0\ \mathrm V$ at $100^\circ\mathrm C$ . If the response is linear, then

m=\frac{2.0-0.4}{100-20}=0.02\ \mathrm{V}/^\circ\mathrm C.

Using $$(20,0.4)$$ ,

\[V-0.4=0.02(T-20),\]

so $$V=0.02T$$ . The zero intercept is a feature of this particular data set, not a general property of sensors.

Subtract first so the intercept disappears. Two points on one non-vertical line share the same intercept, so subtracting their equations leaves only the rate of change. Once that slope is known, point-slope form rebuilds the whole line from either measured point without requiring a separately guessed intercept.

m=\frac{11-2}{4-1}=3,\qquad y-2=3(x-1)

Subtract the y-values and the non-zero x-values in the same order.
Divide the signed rise by the signed run to obtain the slope.
Substitute that slope and either original point into y minus y one equals m times x minus x one.

3.3 Intersections and solution cases

The intersection of $$y=m_1x+b_1$$ and $$y=m_2x+b_2$$ satisfies both equations. Equating them gives

\[m_1x+b_1=m_2x+b_2.\]

If $m_1\ne m_2$ , then

x=\frac{b_2-b_1}{m_1-m_2}

and there is one intersection. If $$m_1=m_2$$ but $b_1\ne b_2$ , the lines are parallel and there is no solution. If both slope and intercept agree, the two equations describe the same line and there are infinitely many solutions.

Three pictures, three algebraic outcomes. Distinct non-parallel lines meet once; distinct parallel lines never share a point; coincident lines share every point. Before manipulating symbols, compare slopes and intercepts: that comparison predicts whether a unique answer is even possible.

One intersection, no intersection, and infinitely many intersections for pairs of lines.

3.4 Quadratic functions and their shape

A quadratic function has the form

f(x)=ax^2+bx+c,\qquad a\ne0.

Its graph is a parabola. The sign of $$a$$ controls opening: $$a>0$$ gives a bowl opening upward; $$a<0$$ gives a bowl opening downward. To see the centre and extreme value clearly, rewrite the quadratic in vertex form.

Start by factoring $$a$$ from the first two terms:

ax^2+bx+c=a\left(x^2+\frac ba x\right)+c.

Inside the bracket, add and subtract $\left(\frac{b}{2a}\right)^2$ :

f(x)=a\left[x^2+\frac ba x+\left(\frac{b}{2a}\right)^2 -\left(\frac{b}{2a}\right)^2\right]+c.

The first three terms form a square, hence

f(x)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.

The vertex is therefore

\left(-\frac b{2a},\ c-\frac{b^2}{4a}\right).

Example — Completing the square

Find the vertex and minimum of $$f(x)=2x^2-8x+3$$ .

\begin{aligned} f(x)&=2(x^2-4x)+3\\ &=2[(x-2)^2-4]+3\\ &=2(x-2)^2-5. \end{aligned}

Because $2(x-2)^2\ge0$ , the minimum is $$-5$$ at $$x=2$$ .

Derivation. Completing the square manufactures the square that reveals a parabola's vertex. The added and subtracted quantity is the same, so the function is not changed; only its geometry becomes visible.

\[x^2-6x+4=(x-3)^2-5\]

A parabola shown with vertex and possible root configurations.

3.5 Roots and the quadratic formula

Roots solve $$ax^2+bx+c=0$$ . Using the completed-square expression,

From the completed-square form, move the constant term carefully:

a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c.

Multiply by $$4a$$ :

4a^2\left(x+\frac{b}{2a}\right)^2=b^2-4ac.

Taking square roots and solving for $$x$$ gives

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

The quantity $\Delta=b^2-4ac$ is the discriminant. If $\Delta>0$ , there are two distinct real roots; if $\Delta=0$ , one repeated real root; and if $\Delta<0$ , no real roots. Complex roots are studied in Chapter 14.

The discriminant is a geometric count. After completing the square, the square root in the quadratic formula is real twice, once, or not at all according as the discriminant is positive, zero, or negative. This is exactly the count of x-axis intersections of a parabola, not a separate graphical rule.

x^2-6x+4=(x-3)^2-5,\qquad \Delta\gtrless0\Longleftrightarrow 2,1,0\text{ real roots}

Compute b squared minus four a c before taking a square root.
Classify its sign as positive, zero, or negative.
Use the matching root count to check the expected number of graph intersections.

3.6 Lines and parabolas as geometric models

Slope is a rate with units. If voltage is plotted against time, slope has units V/s; an intercept has voltage units. A negative slope says the output decreases as input increases. A line through two measurements is an interpolation model, not proof that the same rule remains valid far outside the measurements.

Example — parallel and perpendicular lines

The line $$2x-3y=6$$ is $$y=(2/3)x-2$$ . A parallel line through $$(3,1)$$ is

y-1=\frac23(x-3).

A perpendicular non-vertical line has slope $$-3/2$$ , because the product of slopes is $$-1$$ , so through the same point it is

y-1=-\frac32(x-3).

Vertical and horizontal lines are the special case: one is perpendicular to the other, but a vertical line has no finite slope to invert.

Example — discriminant as a graphical test

For $$2x^2+4x+5=0$$ ,

\Delta=4^2-4(2)(5)=-24.

There are no real roots, so the upward-opening parabola never meets the $$x$$ -axis. Completing the square confirms this:

\[2x^2+4x+5=2(x+1)^2+3>0.\]

Engineering Example — a line is a local approximation

Near an operating current $$I_0$$ , a non-linear device curve can be approximated by $$V(I_0)+m(I-I_0)$$ . The slope $$m$$ is a local resistance-like rate $\Delta V/\Delta I$ . Chapter 7 makes that local slope precise as a derivative.

[Figure: Nonparallel, parallel, and coincident line pairs; an upward parabola with zero, one, and two possible

\(x\)

-axis intersections.]

An intersection makes both models true. Replacing the line's output by the parabola's output produces one quadratic equation in the shared input. Its discriminant predicts whether the line crosses twice, is tangent once, or misses the parabola; substitution then returns each corresponding vertical coordinate.

ax^2+bx+c=mx+d\Longrightarrow ax^2+(b-m)x+(c-d)=0

A line crossing, touching, or missing a parabola according to the resulting quadratic root count.

Problem-Solving Strategy

For a line, choose slope-intercept form when the slope/intercept are known and point-slope form when a point and slope are known. For a quadratic, use vertex form to study shape and extrema, factorisation when factors are visible, and the quadratic formula when a general reliable method is needed.

Exercises

Find the equation of the line through $$(2,5)$$ and $$(6,17)$$ .
Determine whether $$3x-2y=7$$ and $$6x-4y=14$$ have one, none, or infinitely many solutions.
Complete the square for $$x^2+6x-7$$ and state its vertex.
Solve $$3x^2-5x-2=0$$ and verify both roots by substitution.
A power model is $P(R)=\frac{36R}{(R+3)^2}$ watts for $$R>0$$ . Sketch its qualitative behaviour and identify why it is not a linear model.
Find a line through $$(1,4)$$ perpendicular to $$3x+2y=7$$ .
Use discriminant and vertex form to explain why $$x^2-6x+13=0$$ has no real root.
A calibration line passes through $$(10,0.8)$$ and $$(70,2.6)$$ . Find its model and state the units of its slope.