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Week 3: Quadratic Functions, Graphs & Equations

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Week 3: Quadratic Functions, Graphs & Equations

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2026-09-12

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Mathematics I — Week 3: Quadratic Functions, Graphs & Equations

IITM’s Week 3 covers quadratic functions, minima/maxima, vertex and slope-related interpretation, and quadratic equations.

TL;DR

A quadratic function has the form

f(x)=ax^2+bx+c,\qquad a\neq 0.

Its graph is a parabola.

The most important Week 3 ideas are:

standard form, factored form, vertex form
roots / $x$ -intercepts
$y$ -intercept
axis of symmetry
vertex
minimum or maximum
discriminant
quadratic formula
relation between roots and coefficients
how changing $a,b,c$ changes the graph

1. What is a quadratic function?

A quadratic function is:

\boxed{ f(x)=ax^2+bx+c }

where:

a,b,c\in\mathbb R

and:

a\neq0.

Why must

a\neq0

Because if:

a=0,

then:

f(x)=bx+c,

which is linear, not quadratic.

2. Why "quadratic"?

The word comes from the Latin quadratus, meaning "square."

That is because the highest power of the variable is:

x^2.

So:

3x^2+2x+1

is quadratic.

But:

x^3+2x

is cubic.

3. Degree

For:

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

the degree is:

2.

The degree is the highest exponent of

x

with a nonzero coefficient.

4. Graph of a quadratic

The graph of a quadratic function is a parabola.

Example:

y=x^2

looks roughly like:


text
      y
      ↑
      |
   *     *
    *   *
     * *
      *
------+------→ x

It is symmetric around the

y

-axis.

5. Direction of opening

The coefficient

a

determines whether the parabola opens upward or downward.

If $a>0$

The parabola opens upward:

\cup

It has a minimum.

If $a<0$

The parabola opens downward:

\cap

It has a maximum.

This is one of the fastest graph facts to identify.

6. Effect of magnitude of $a$

Compare:

y=x^2

y=3x^2

y=\frac12x^2.

Larger:

|a|

makes the parabola appear narrower/steeper.

Smaller:

|a|

makes it appear wider/flatter.

7. Standard form

The standard form is:

\boxed{ f(x)=ax^2+bx+c }

This form is especially useful for:

identifying coefficients
using the quadratic formula
finding the discriminant
getting the $y$ -intercept immediately

At:

x=0,

we get:

f(0)=c.

Therefore the

y

-intercept is:

\boxed{(0,c)}.

8. Roots / zeros / $x$ -intercepts

A root or zero is a value of

x

satisfying:

f(x)=0.

For:

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

roots satisfy:

ax^2+bx+c=0.

Graphically, roots are where the parabola intersects the

x

-axis.

9. Example

Consider:

f(x)=x^2-5x+6.

Set:

f(x)=0.

Then:

x^2-5x+6=0.

Factor:

(x-2)(x-3)=0.

Therefore:

x=2

or:

x=3.

So the

x

-intercepts are:

(2,0),\qquad(3,0).

10. Factored form

If roots are:

r_1,\qquad r_2,

then a quadratic can be written as:

\boxed{ f(x)=a(x-r_1)(x-r_2) }

This form makes the roots immediately visible.

Example:

f(x)=2(x-1)(x+4)

has roots:

1

and:

-4.

11. Axis of symmetry

Every parabola has a vertical line through its center called the axis of symmetry.

For:

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

the axis is:

\boxed{ x=-\frac{b}{2a} }

This is one of the most important Week 3 formulas.

12. Vertex

The vertex is the turning point of the parabola.

Its

x

-coordinate is:

x_v=-\frac{b}{2a}.

Then:

y_v=f\left(-\frac{b}{2a}\right).

Therefore:

\boxed{ V= \left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) }

13. Minimum or maximum

If:

a>0,

the vertex is the minimum point.

If:

a<0,

the vertex is the maximum point.

So once you know the vertex and the sign of

a

, you know the extremum.

14. Example: vertex

Consider:

f(x)=x^2-6x+5.

Here:

a=1,\qquad b=-6.

Thus:

x_v= -\frac{-6}{2(1)} = 3.

Now:

f(3)=9-18+5=-4.

Therefore:

\boxed{V=(3,-4)}.

Since:

a>0,

the parabola opens upward.

So:

\boxed{\text{minimum value}=-4}.

15. Vertex form

A quadratic can also be written as:

\boxed{ f(x)=a(x-h)^2+k }

This is called vertex form.

The vertex is immediately:

\boxed{(h,k)}.

The axis of symmetry is:

\boxed{x=h}.

16. Example

Consider:

f(x)=2(x-3)^2+5.

Then:

h=3,\qquad k=5.

Therefore vertex:

(3,5).

Since:

a=2>0,

it opens upward.

Hence minimum value:

5.

17. Completing the square

Completing the square converts standard form into vertex form.

Start with:

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

Factor

a

from the first two terms:

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

Now add and subtract:

\left(\frac{b}{2a}\right)^2.

Then:

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

So:

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

Thus the vertex is:

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

18. Alternative expression for vertex $y$ -coordinate

Since:

\Delta=b^2-4ac,

we can write:

c-\frac{b^2}{4a} = -\frac{b^2-4ac}{4a} = -\frac{\Delta}{4a}.

Therefore:

\boxed{ V= \left( -\frac{b}{2a}, -\frac{\Delta}{4a} \right) }

This is a useful connection between the graph and discriminant.

19. Quadratic equation

A quadratic equation is:

\boxed{ ax^2+bx+c=0 }

with:

a\neq0.

The goal is to find values of

x

satisfying the equation.

20. Solving by factorization

Example:

x^2-7x+12=0.

We want two numbers whose:

product is $12$
sum is $-7$

Those numbers are:

-3,\quad-4.

So:

(x-3)(x-4)=0.

Hence:

x=3

or:

x=4.

21. Zero-product principle

The reason factorization works is:

AB=0

implies:

A=0

or:

B=0.

Thus:

(x-3)(x-4)=0

means:

x-3=0

or:

x-4=0.

22. Quadratic formula

For:

ax^2+bx+c=0,

the roots are:

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

The expression:

b^2-4ac

is called the discriminant.

23. Discriminant

Define:

\boxed{ \Delta=b^2-4ac }

It tells us the nature of the roots.

24. Case 1: $\Delta>0$

If:

\Delta>0,

there are two distinct real roots.

Graphically, the parabola crosses the

x

-axis at two distinct points.


text
\        /
 \      /
--\----/------ x
   \  /
    \/

25. Case 2: $\Delta=0$

If:

\Delta=0,

there is one repeated real root.

The two roots are equal:

x= -\frac b{2a}.

Graphically, the parabola just touches the

x

-axis at its vertex.

26. Case 3: $\Delta<0$

If:

\Delta<0,

there are no real roots.

Graphically, the parabola never intersects the

x

-axis.

Over complex numbers, there are still two complex roots.

But for real-coordinate graph interpretation:

\boxed{\text{no real }x\text{-intercepts}}

27. Root/discriminant cheat sheet

Discriminant	Roots	Graph
$\Delta>0$	two distinct real	crosses $x$ -axis twice
$\Delta=0$	one repeated real	touches $x$ -axis
$\Delta<0$	no real roots	no $x$ -axis intersection

28. Deriving the quadratic formula

This is worth understanding once instead of memorizing blindly.

Start:

ax^2+bx+c=0.

Divide by

a

x^2+\frac ba x+\frac ca=0.

Move the constant:

x^2+\frac ba x = -\frac ca.

Complete the square:

x^2+\frac ba x + \frac{b^2}{4a^2} = -\frac ca + \frac{b^2}{4a^2}.

Then:

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

Take square roots:

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

Thus:

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

29. Sum and product of roots

Suppose the roots are:

\alpha,\beta.

Then:

a(x-\alpha)(x-\beta) = ax^2-a(\alpha+\beta)x+a\alpha\beta.

Compare this with:

ax^2+bx+c.

Therefore:

\boxed{ \alpha+\beta=-\frac ba }

and:

\boxed{ \alpha\beta=\frac ca }

These are extremely useful.

30. Example

For:

2x^2-7x+3=0,

we have:

a=2,\quad b=-7,\quad c=3.

Therefore sum of roots:

\alpha+\beta = -\frac{-7}{2} = \frac72.

Product:

\alpha\beta = \frac32.

Without even solving the equation.

31. Constructing a quadratic from roots

If roots are:

\alpha,\beta,

then:

(x-\alpha)(x-\beta)=0

gives:

x^2-(\alpha+\beta)x+\alpha\beta=0.

Example:

Roots:

2,\quad5.

Then:

(x-2)(x-5)=0

so:

x^2-7x+10=0.

32. Repeated root

If both roots equal:

r,

then:

f(x)=a(x-r)^2.

This is exactly why the parabola touches the

x

-axis at one point.

That point is also the vertex.

33. Relationship between roots and axis of symmetry

If roots are:

r_1,\qquad r_2,

then the axis of symmetry lies exactly halfway between them:

\boxed{ x= \frac{r_1+r_2}{2} }

But:

r_1+r_2=-\frac ba.

Therefore:

x= -\frac b{2a}.

So the familiar vertex formula is also just:

the midpoint of the roots.

Very useful intuition.

34. Example

Suppose roots are:

2,\qquad8.

Then axis:

x= \frac{2+8}{2} = 5.

The parabola is symmetric around:

x=5.

35. Domain of a quadratic

For the ordinary quadratic:

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

there are no restrictions on

x

Therefore:

\boxed{ \text{Domain}=\mathbb R }

36. Range of a quadratic

The range depends on the vertex.

If:

a>0,

the vertex gives the minimum:

y_{\min}=k.

Then:

\boxed{ \text{Range}=[k,\infty) }

for vertex:

(h,k).

If:

a<0,

the vertex gives the maximum:

\boxed{ \text{Range}=(-\infty,k] }

37. Example

For:

f(x)=(x-2)^2-5,

vertex:

(2,-5).

Since:

a=1>0,

minimum:

-5.

Thus:

\boxed{ \text{Range}=[-5,\infty) }

38. Transformations from $y=x^2$

The base parabola is:

y=x^2.

Now consider:

y=(x-h)^2+k.

This shifts the graph:

right by $h$ if $h>0$
left by $|h|$ if $h<0$
up by $k$ if $k>0$
down by $|k|$ if $k<0$

39. The sign trap

For:

y=(x-3)^2,

the graph shifts right by 3, not left.

Why?

The vertex occurs when:

x-3=0,

so:

x=3.

Similarly:

y=(x+4)^2

has vertex at:

x=-4.

The sign inside the bracket appears reversed.

40. Reflection

If:

y=f(x),

then:

y=-f(x)

reflects the graph across the

x

-axis.

Thus:

y=x^2

becomes:

y=-x^2.

Upward-opening becomes downward-opening.

41. Vertical stretching/compression

For:

y=ax^2:

$|a|>1$ gives vertical stretching
$0<|a|<1$ gives vertical compression

Visually:

stretched → narrower
compressed → wider

42. Finding a quadratic from three points

A quadratic has three unknown coefficients:

a,b,c.

So three suitable points usually determine it uniquely.

Suppose it passes through:

(0,1),\quad(1,4),\quad(2,9).

Let:

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

From

(0,1)

c=1.

From

(1,4)

a+b+1=4

so:

a+b=3.

From

(2,9)

4a+2b+1=9

so:

2a+b=4.

Subtract:

(2a+b)-(a+b)=4-3

thus:

a=1.

Then:

b=2.

Therefore:

\boxed{ f(x)=x^2+2x+1 }

which is:

(x+1)^2.

43. Slope of a quadratic is not constant

For a straight line:

y=mx+c,

slope is constant.

For a quadratic:

y=ax^2+bx+c,

the steepness changes with

x

This is why the graph curves.

Later, calculus formalizes this using the derivative:

f'(x)=2ax+b.

You do not need derivatives yet to understand the idea:

the slope changes continuously as you move along the parabola.

44. What happens at the vertex?

At the vertex:

decreasing switches to increasing, or
increasing switches to decreasing.

So the tangent there becomes horizontal.

In later calculus language:

\text{slope}=0.

Indeed:

2ax+b=0

gives:

x=-\frac b{2a}.

Exactly the vertex formula.

This is a nice preview of Week 8 derivatives.

45. Increasing and decreasing behavior

For:

a>0,

the parabola decreases until:

x=-\frac b{2a},

then increases afterward.

So:

f(x)\text{ decreases on } \left(-\infty,-\frac b{2a}\right)

and increases on:

\left(-\frac b{2a},\infty\right).

For:

a<0,

the behavior reverses.

46. Sign of a quadratic from its graph

Suppose:

a>0

and there are two roots:

r_1<r_2.

Then:

f(x)>0

outside the roots:

x<r_1

or:

x>r_2.

And:

f(x)<0

between them:

r_1<x<r_2.

For a downward-opening parabola, signs reverse.

This becomes useful for quadratic inequalities.

47. Example

Consider:

f(x)=(x-2)(x-5).

Roots:

2,\quad5.

Since:

a=1>0,

the graph opens upward.

Therefore:

f(x)>0

for:

x<2

or:

x>5.

And:

f(x)<0

for:

2<x<5.

48. Quadratic inequalities

Suppose:

x^2-5x+6>0.

Factor:

(x-2)(x-3)>0.

Critical points:

2,\quad3.

Since the parabola opens upward:

f(x)>0

outside the roots.

Thus:

\boxed{ x<2\quad\text{or}\quad x>3 }

Common trap: root vs vertex

Roots are points where:

y=0.

The vertex is the turning point.

They are different concepts.

Example:

f(x)=x^2-4.

Roots:

x=\pm2.

Vertex:

(0,-4).

Common trap: vertex

x

-coordinate

For:

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

the vertex is not:

-\frac ba.

It is:

\boxed{ -\frac b{2a} }

The factor

2

matters.

Common trap: discriminant

For:

ax^2+bx+c=0,

the discriminant is:

\boxed{ b^2-4ac }

not:

b^2+4ac.

Common trap:

\pm

Quadratic formula:

x= \frac{-b\pm\sqrt{\Delta}}{2a}.

The

\pm

produces two possible roots:

x_1= \frac{-b+\sqrt\Delta}{2a}

and:

x_2= \frac{-b-\sqrt\Delta}{2a}.

Unless:

\Delta=0,

in which case both coincide.

Common trap:

(-b)^2

-b^2

These are different:

(-b)^2=b^2

but:

-b^2=-(b^2).

For example:

(-3)^2=9

while:

-3^2=-9.

Parentheses matter.

Common trap: interpreting

c

In:

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

c

gives the $y$ -intercept, not necessarily an

x

-intercept.

Because:

f(0)=c.

55. Three useful forms of the same quadratic

Suppose:

f(x)=x^2-6x+5.

Standard form

x^2-6x+5

Useful for:

coefficients
discriminant
quadratic formula

Factored form

(x-1)(x-5)

Useful for:

roots

Vertex form

(x-3)^2-4

Useful for:

vertex
minimum
axis of symmetry

Each representation exposes different information.

This is an important general mathematical pattern:

Same object, different representation, different information becomes obvious.

56. Full graph anatomy example

Take:

f(x)=x^2-4x+3.

Step 1: opening

a=1>0

so it opens upward.

Step 2: $y$ -intercept

c=3

so:

(0,3).

Step 3: roots

x^2-4x+3 = (x-1)(x-3).

So:

x=1,3.

Step 4: axis

x= -\frac{-4}{2} = 2.

Step 5: vertex

f(2)=4-8+3=-1.

Therefore:

V=(2,-1).

Step 6: range

Since it opens upward:

y\geq -1.

So:

\boxed{ [-1,\infty) }

That is essentially a complete quadratic analysis.

57. Fast exam workflow

Given:

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

a good order is:


text
1. Read a,b,c
2. Check sign of a
3. Find y-intercept = c
4. Calculate Δ = b² - 4ac
5. Find roots if needed
6. Find axis x = -b/(2a)
7. Find vertex
8. Infer min/max and range
9. Sketch

This minimizes duplicated work.

58. Connections to previous weeks

Week 1 → functions

A quadratic is a particular function:

f:\mathbb R\to\mathbb R

usually written:

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

Week 2 → coordinate geometry

The graph consists of points:

(x,y)

satisfying:

y=ax^2+bx+c.

Roots are simply intersections with the

x

-axis.

The

y

-intercept is the intersection with the

y

-axis.

So Week 3 builds directly on Week 2.

59. One deeper mental model

Think of:

y=ax^2+bx+c

as producing an output

y

for every input

x

The graph is simply a picture of all input-output pairs:

\{(x,f(x)):x\in\mathbb R\}.

That idea applies to every function you'll study later.

60. Formula sheet

Standard form

\boxed{ f(x)=ax^2+bx+c }

Axis of symmetry

\boxed{ x=-\frac b{2a} }

Vertex

\boxed{ \left( -\frac b{2a}, f\left(-\frac b{2a}\right) \right) }

or:

\boxed{ \left( -\frac b{2a}, -\frac{\Delta}{4a} \right) }

Vertex form

\boxed{ f(x)=a(x-h)^2+k }

Quadratic formula

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

Discriminant

\boxed{ \Delta=b^2-4ac }

Sum of roots

\boxed{ \alpha+\beta=-\frac ba }

Product of roots

\boxed{ \alpha\beta=\frac ca }

61. What you should actually master

By the end of Week 3, you should be able to:

Recognize a quadratic function.
Identify $a,b,c$ correctly.
Determine whether its parabola opens up or down.
Find the $y$ -intercept.
Find roots by factorization.
Use the quadratic formula.
Calculate and interpret the discriminant.
Find the axis of symmetry.
Find the vertex.
Determine minimum/maximum value.
Find the range.
Convert between standard, factored and vertex forms.
Use root sum/product relations.
Sketch a parabola from algebraic information.
Solve simple quadratic inequalities.

If those are comfortable, Week 3 is under control.

62. Concept map


text
Quadratic
f(x)=ax²+bx+c
│
├── coefficient a
│   ├── a>0 → minimum
│   └── a<0 → maximum
│
├── intercepts
│   ├── y-intercept → c
│   └── x-intercepts → roots
│
├── discriminant
│   ├── Δ>0 → 2 real roots
│   ├── Δ=0 → repeated root
│   └── Δ<0 → no real roots
│
├── symmetry
│   └── x=-b/(2a)
│
├── vertex
│
└── representations
    ├── standard
    ├── factored
    └── vertex

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:

Question :: Answer.

Vertex $x$ -coordinate for $y = ax^2 + bx + c$ :: $x = -\frac{b}{2a}$
Discriminant formula for $ax^2 + bx + c = 0$ :: $D = b^2 - 4ac$
Roots nature when discriminant $D > 0$ :: Two distinct real roots
Roots nature when discriminant $D = 0$ :: One real repeated root (tangent to axis)
Roots nature when discriminant $D < 0$ :: Two complex conjugate roots (no x-intercepts)
Vieta's formula for sum of roots :: $\alpha + \beta = -\frac{b}{a}$
Vieta's formula for product of roots :: $\alpha \beta = \frac{c}{a}$
Parabola opening direction if $a > 0$ :: Upward (has a global minimum at vertex)
Parabola opening direction if $a < 0$ :: Downward (has a global maximum at vertex)

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.

Question 1

What is the

x

-coordinate of the vertex of the parabola

f(x) = 2x^2 - 8x + 11

NAT

3 Unit Assessment

Question 2

\alpha

and

\beta

are the roots of

2x^2 - 6x + 5 = 0

, what is the value of

\alpha + \beta

MCQ

3 Unit Assessment

A. 3

B. -3

C. 5/2

D. -5/2