Quiz 3

Week 3: Quadratic Functions, Graphs & Equations

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2026-09-12T00:00:00.000Z
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Parabolas, vertex, axis of symmetry, discriminant, nature of roots, Vieta's relations, and quadratic optimization. # 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**.

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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)=ax2+bx+c,a≠0.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 / xx-intercepts
  • yy-intercept
  • axis of symmetry
  • vertex
  • minimum or maximum
  • discriminant
  • quadratic formula
  • relation between roots and coefficients
  • how changing a,b,ca,b,c changes the graph

1. What is a quadratic function?

A quadratic function is:
f(x)=ax2+bx+c\boxed{ f(x)=ax^2+bx+c }
where:
a,b,c∈Ra,b,c\in\mathbb R
and:
a≠0.a\neq0.
Why must a≠0a\neq0?
Because if:
a=0,a=0,
then:
f(x)=bx+c,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:
x2.x^2.
So:
3x2+2x+13x^2+2x+1
is quadratic.
But:
x3+2xx^3+2x
is cubic.

3. Degree

For:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
the degree is:
2.2.
The degree is the highest exponent of xx with a nonzero coefficient.

4. Graph of a quadratic

The graph of a quadratic function is a parabola.
Example:
y=x2y=x^2
looks roughly like:
text
      y
      ↑
      |
   *     *
    *   *
     * *
      *
------+------→ x
It is symmetric around the yy-axis.

5. Direction of opening

The coefficient aa determines whether the parabola opens upward or downward.

If a>0a>0

The parabola opens upward:
∪\cup
It has a minimum.

If a<0a<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 aa

Compare:
y=x2y=x^2 y=3x2y=3x^2 y=12x2.y=\frac12x^2.
Larger:
∣a∣|a|
makes the parabola appear narrower/steeper.
Smaller:
∣a∣|a|
makes it appear wider/flatter.

7. Standard form

The standard form is:
f(x)=ax2+bx+c\boxed{ f(x)=ax^2+bx+c }
This form is especially useful for:
  • identifying coefficients
  • using the quadratic formula
  • finding the discriminant
  • getting the yy-intercept immediately
At:
x=0,x=0,
we get:
f(0)=c.f(0)=c.
Therefore the yy-intercept is:
(0,c).\boxed{(0,c)}.

8. Roots / zeros / xx-intercepts

A root or zero is a value of xx satisfying:
f(x)=0.f(x)=0.
For:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
roots satisfy:
ax2+bx+c=0.ax^2+bx+c=0.
Graphically, roots are where the parabola intersects the xx-axis.

9. Example

Consider:
f(x)=x2−5x+6.f(x)=x^2-5x+6.
Set:
f(x)=0.f(x)=0.
Then:
x2−5x+6=0.x^2-5x+6=0.
Factor:
(x−2)(x−3)=0.(x-2)(x-3)=0.
Therefore:
x=2x=2
or:
x=3.x=3.
So the xx-intercepts are:
(2,0),(3,0).(2,0),\qquad(3,0).

10. Factored form

If roots are:
r1,r2,r_1,\qquad r_2,
then a quadratic can be written as:
f(x)=a(x−r1)(x−r2)\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)f(x)=2(x-1)(x+4)
has roots:
11
and:
−4.-4.

11. Axis of symmetry

Every parabola has a vertical line through its center called the axis of symmetry.
For:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
the axis is:
x=−b2a\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 xx-coordinate is:
xv=−b2a.x_v=-\frac{b}{2a}.
Then:
yv=f(−b2a).y_v=f\left(-\frac{b}{2a}\right).
Therefore:
V=(−b2a,f(−b2a))\boxed{ V= \left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) }

13. Minimum or maximum

If:
a>0,a>0,
the vertex is the minimum point.
If:
a<0,a<0,
the vertex is the maximum point.
So once you know the vertex and the sign of aa, you know the extremum.

14. Example: vertex

Consider:
f(x)=x2−6x+5.f(x)=x^2-6x+5.
Here:
a=1,b=−6.a=1,\qquad b=-6.
Thus:
xv=−−62(1)=3.x_v= -\frac{-6}{2(1)} = 3.
Now:
f(3)=9−18+5=−4.f(3)=9-18+5=-4.
Therefore:
V=(3,−4).\boxed{V=(3,-4)}.
Since:
a>0,a>0,
the parabola opens upward.
So:
minimum value=−4.\boxed{\text{minimum value}=-4}.

15. Vertex form

A quadratic can also be written as:
f(x)=a(x−h)2+k\boxed{ f(x)=a(x-h)^2+k }
This is called vertex form.
The vertex is immediately:
(h,k).\boxed{(h,k)}.
The axis of symmetry is:
x=h.\boxed{x=h}.

16. Example

Consider:
f(x)=2(x−3)2+5.f(x)=2(x-3)^2+5.
Then:
h=3,k=5.h=3,\qquad k=5.
Therefore vertex:
(3,5).(3,5).
Since:
a=2>0,a=2>0,
it opens upward.
Hence minimum value:
5.5.

17. Completing the square

Completing the square converts standard form into vertex form.
Start with:
f(x)=ax2+bx+c.f(x)=ax^2+bx+c.
Factor aa from the first two terms:
f(x)=a(x2+bax)+c.f(x) = a\left( x^2+\frac ba x \right)+c.
Now add and subtract:
(b2a)2.\left(\frac{b}{2a}\right)^2.
Then:
f(x)=a[(x+b2a)2−(b2a)2]+c.f(x) = a\left[ \left( x+\frac{b}{2a} \right)^2 - \left( \frac{b}{2a} \right)^2 \right]+c.
So:
f(x)=a(x+b2a)2+c−b24a.f(x)= a\left( x+\frac{b}{2a} \right)^2 + c-\frac{b^2}{4a}.
Thus the vertex is:
(−b2a,c−b24a)\boxed{ \left( -\frac{b}{2a}, c-\frac{b^2}{4a} \right) }

18. Alternative expression for vertex yy-coordinate

Since:
Δ=b2−4ac,\Delta=b^2-4ac,
we can write:
c−b24a=−b2−4ac4a=−Δ4a.c-\frac{b^2}{4a} = -\frac{b^2-4ac}{4a} = -\frac{\Delta}{4a}.
Therefore:
V=(−b2a,−Δ4a)\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:
ax2+bx+c=0\boxed{ ax^2+bx+c=0 }
with:
a≠0.a\neq0.
The goal is to find values of xx satisfying the equation.

20. Solving by factorization

Example:
x2−7x+12=0.x^2-7x+12=0.
We want two numbers whose:
  • product is 1212
  • sum is −7-7
Those numbers are:
−3,−4.-3,\quad-4.
So:
(x−3)(x−4)=0.(x-3)(x-4)=0.
Hence:
x=3x=3
or:
x=4.x=4.

21. Zero-product principle

The reason factorization works is:
AB=0AB=0
implies:
A=0A=0
or:
B=0.B=0.
Thus:
(x−3)(x−4)=0(x-3)(x-4)=0
means:
x−3=0x-3=0
or:
x−4=0.x-4=0.

22. Quadratic formula

For:
ax2+bx+c=0,ax^2+bx+c=0,
the roots are:
x=−b±b2−4ac2a\boxed{ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} }
The expression:
b2−4acb^2-4ac
is called the discriminant.

23. Discriminant

Define:
Δ=b2−4ac\boxed{ \Delta=b^2-4ac }
It tells us the nature of the roots.

24. Case 1: Δ>0\Delta>0

If:
Δ>0,\Delta>0,
there are two distinct real roots.
Graphically, the parabola crosses the xx-axis at two distinct points.
text
\        /
 \      /
--\----/------ x
   \  /
    \/

25. Case 2: Δ=0\Delta=0

If:
Δ=0,\Delta=0,
there is one repeated real root.
The two roots are equal:
x=−b2a.x= -\frac b{2a}.
Graphically, the parabola just touches the xx-axis at its vertex.

26. Case 3: Δ<0\Delta<0

If:
Δ<0,\Delta<0,
there are no real roots.
Graphically, the parabola never intersects the xx-axis.
Over complex numbers, there are still two complex roots.
But for real-coordinate graph interpretation:
no real x-intercepts\boxed{\text{no real }x\text{-intercepts}}

27. Root/discriminant cheat sheet

DiscriminantRootsGraph
Δ>0\Delta>0two distinct realcrosses xx-axis twice
Δ=0\Delta=0one repeated realtouches xx-axis
Δ<0\Delta<0no real rootsno xx-axis intersection

28. Deriving the quadratic formula

This is worth understanding once instead of memorizing blindly.
Start:
ax2+bx+c=0.ax^2+bx+c=0.
Divide by aa:
x2+bax+ca=0.x^2+\frac ba x+\frac ca=0.
Move the constant:
x2+bax=−ca.x^2+\frac ba x = -\frac ca.
Complete the square:
x2+bax+b24a2=−ca+b24a2.x^2+\frac ba x + \frac{b^2}{4a^2} = -\frac ca + \frac{b^2}{4a^2}.
Then:
(x+b2a)2=b2−4ac4a2.\left( x+\frac b{2a} \right)^2 = \frac{b^2-4ac}{4a^2}.
Take square roots:
x+b2a=±b2−4ac2a.x+\frac b{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}.
Thus:
x=−b±b2−4ac2a.x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}.

29. Sum and product of roots

Suppose the roots are:
α,β.\alpha,\beta.
Then:
a(x−α)(x−β)=ax2−a(α+β)x+aαβ.a(x-\alpha)(x-\beta) = ax^2-a(\alpha+\beta)x+a\alpha\beta.
Compare this with:
ax2+bx+c.ax^2+bx+c.
Therefore:
α+β=−ba\boxed{ \alpha+\beta=-\frac ba }
and:
αβ=ca\boxed{ \alpha\beta=\frac ca }
These are extremely useful.

30. Example

For:
2x2−7x+3=0,2x^2-7x+3=0,
we have:
a=2,b=−7,c=3.a=2,\quad b=-7,\quad c=3.
Therefore sum of roots:
α+β=−−72=72.\alpha+\beta = -\frac{-7}{2} = \frac72.
Product:
αβ=32.\alpha\beta = \frac32.
Without even solving the equation.

31. Constructing a quadratic from roots

If roots are:
α,β,\alpha,\beta,
then:
(x−α)(x−β)=0(x-\alpha)(x-\beta)=0
gives:
x2−(α+β)x+αβ=0.x^2-(\alpha+\beta)x+\alpha\beta=0.
Example:
Roots:
2,5.2,\quad5.
Then:
(x−2)(x−5)=0(x-2)(x-5)=0
so:
x2−7x+10=0.x^2-7x+10=0.

32. Repeated root

If both roots equal:
r,r,
then:
f(x)=a(x−r)2.f(x)=a(x-r)^2.
This is exactly why the parabola touches the xx-axis at one point.
That point is also the vertex.

33. Relationship between roots and axis of symmetry

If roots are:
r1,r2,r_1,\qquad r_2,
then the axis of symmetry lies exactly halfway between them:
x=r1+r22\boxed{ x= \frac{r_1+r_2}{2} }
But:
r1+r2=−ba.r_1+r_2=-\frac ba.
Therefore:
x=−b2a.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:
2,8.2,\qquad8.
Then axis:
x=2+82=5.x= \frac{2+8}{2} = 5.
The parabola is symmetric around:
x=5.x=5.

35. Domain of a quadratic

For the ordinary quadratic:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
there are no restrictions on xx.
Therefore:
Domain=R\boxed{ \text{Domain}=\mathbb R }

36. Range of a quadratic

The range depends on the vertex.
If:
a>0,a>0,
the vertex gives the minimum:
ymin⁡=k.y_{\min}=k.
Then:
Range=[k,∞)\boxed{ \text{Range}=[k,\infty) }
for vertex:
(h,k).(h,k).
If:
a<0,a<0,
the vertex gives the maximum:
Range=(−∞,k]\boxed{ \text{Range}=(-\infty,k] }

37. Example

For:
f(x)=(x−2)2−5,f(x)=(x-2)^2-5,
vertex:
(2,−5).(2,-5).
Since:
a=1>0,a=1>0,
minimum:
−5.-5.
Thus:
Range=[−5,∞)\boxed{ \text{Range}=[-5,\infty) }

38. Transformations from y=x2y=x^2

The base parabola is:
y=x2.y=x^2.
Now consider:
y=(x−h)2+k.y=(x-h)^2+k.
This shifts the graph:
  • right by hh if h>0h>0
  • left by ∣h∣|h| if h<0h<0
  • up by kk if k>0k>0
  • down by ∣k∣|k| if k<0k<0

39. The sign trap

For:
y=(x−3)2,y=(x-3)^2,
the graph shifts right by 3, not left.
Why?
The vertex occurs when:
x−3=0,x-3=0,
so:
x=3.x=3.
Similarly:
y=(x+4)2y=(x+4)^2
has vertex at:
x=−4.x=-4.
The sign inside the bracket appears reversed.

40. Reflection

If:
y=f(x),y=f(x),
then:
y=−f(x)y=-f(x)
reflects the graph across the xx-axis.
Thus:
y=x2y=x^2
becomes:
y=−x2.y=-x^2.
Upward-opening becomes downward-opening.

41. Vertical stretching/compression

For:
y=ax2:y=ax^2:
  • ∣a∣>1|a|>1 gives vertical stretching
  • 0<∣a∣<10<|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.a,b,c.
So three suitable points usually determine it uniquely.
Suppose it passes through:
(0,1),(1,4),(2,9).(0,1),\quad(1,4),\quad(2,9).
Let:
f(x)=ax2+bx+c.f(x)=ax^2+bx+c.
From (0,1)(0,1):
c=1.c=1.
From (1,4)(1,4):
a+b+1=4a+b+1=4
so:
a+b=3.a+b=3.
From (2,9)(2,9):
4a+2b+1=94a+2b+1=9
so:
2a+b=4.2a+b=4.
Subtract:
(2a+b)−(a+b)=4−3(2a+b)-(a+b)=4-3
thus:
a=1.a=1.
Then:
b=2.b=2.
Therefore:
f(x)=x2+2x+1\boxed{ f(x)=x^2+2x+1 }
which is:
(x+1)2.(x+1)^2.

43. Slope of a quadratic is not constant

For a straight line:
y=mx+c,y=mx+c,
slope is constant.
For a quadratic:
y=ax2+bx+c,y=ax^2+bx+c,
the steepness changes with xx.
This is why the graph curves.
Later, calculus formalizes this using the derivative:
f′(x)=2ax+b.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:
slope=0.\text{slope}=0.
Indeed:
2ax+b=02ax+b=0
gives:
x=−b2a.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,a>0,
the parabola decreases until:
x=−b2a,x=-\frac b{2a},
then increases afterward.
So:
f(x) decreases on (−∞,−b2a)f(x)\text{ decreases on } \left(-\infty,-\frac b{2a}\right)
and increases on:
(−b2a,∞).\left(-\frac b{2a},\infty\right).
For:
a<0,a<0,
the behavior reverses.

46. Sign of a quadratic from its graph

Suppose:
a>0a>0
and there are two roots:
r1<r2.r_1<r_2.
Then:
f(x)>0f(x)>0
outside the roots:
x<r1x<r_1
or:
x>r2.x>r_2.
And:
f(x)<0f(x)<0
between them:
r1<x<r2.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).f(x)=(x-2)(x-5).
Roots:
2,5.2,\quad5.
Since:
a=1>0,a=1>0,
the graph opens upward.
Therefore:
f(x)>0f(x)>0
for:
x<2x<2
or:
x>5.x>5.
And:
f(x)<0f(x)<0
for:
2<x<5.2<x<5.

48. Quadratic inequalities

Suppose:
x2−5x+6>0.x^2-5x+6>0.
Factor:
(x−2)(x−3)>0.(x-2)(x-3)>0.
Critical points:
2,3.2,\quad3.
Since the parabola opens upward:
f(x)>0f(x)>0
outside the roots.
Thus:
x<2orx>3\boxed{ x<2\quad\text{or}\quad x>3 }
Common trap: root vs vertex
Roots are points where:
y=0.y=0.
The vertex is the turning point.
They are different concepts.
Example:
f(x)=x2−4.f(x)=x^2-4.
Roots:
x=±2.x=\pm2.
Vertex:
(0,−4).(0,-4).
Common trap: vertex
xx-coordinate
For:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
the vertex is not:
−ba.-\frac ba.
It is:
−b2a\boxed{ -\frac b{2a} }
The factor 22 matters.
Common trap: discriminant
For:
ax2+bx+c=0,ax^2+bx+c=0,
the discriminant is:
b2−4ac\boxed{ b^2-4ac }
not:
b2+4ac.b^2+4ac.
Common trap:
±\pm
Quadratic formula:
x=−b±Δ2a.x= \frac{-b\pm\sqrt{\Delta}}{2a}.
The ±\pm produces two possible roots:
x1=−b+Δ2ax_1= \frac{-b+\sqrt\Delta}{2a}
and:
x2=−b−Δ2a.x_2= \frac{-b-\sqrt\Delta}{2a}.
Unless:
Δ=0,\Delta=0,
in which case both coincide.
Common trap:
(−b)2(-b)^2 vs −b2-b^2
These are different:
(−b)2=b2(-b)^2=b^2
but:
−b2=−(b2).-b^2=-(b^2).
For example:
(−3)2=9(-3)^2=9
while:
−32=−9.-3^2=-9.
Parentheses matter.
Common trap: interpreting
cc
In:
f(x)=ax2+bx+c,f(x)=ax^2+bx+c,
cc gives the yy-intercept, not necessarily an xx-intercept.
Because:
f(0)=c.f(0)=c.

55. Three useful forms of the same quadratic

Suppose:
f(x)=x2−6x+5.f(x)=x^2-6x+5.

Standard form

x2−6x+5x^2-6x+5
Useful for:
  • coefficients
  • discriminant
  • quadratic formula

Factored form

(x−1)(x−5)(x-1)(x-5)
Useful for:
  • roots

Vertex form

(x−3)2−4(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)=x2−4x+3.f(x)=x^2-4x+3.

Step 1: opening

a=1>0a=1>0
so it opens upward.

Step 2: yy-intercept

c=3c=3
so:
(0,3).(0,3).

Step 3: roots

x2−4x+3=(x−1)(x−3).x^2-4x+3 = (x-1)(x-3).
So:
x=1,3.x=1,3.

Step 4: axis

x=−−42=2.x= -\frac{-4}{2} = 2.

Step 5: vertex

f(2)=4−8+3=−1.f(2)=4-8+3=-1.
Therefore:
V=(2,−1).V=(2,-1).

Step 6: range

Since it opens upward:
y≥−1.y\geq -1.
So:
[−1,∞)\boxed{ [-1,\infty) }
That is essentially a complete quadratic analysis.

57. Fast exam workflow

Given:
f(x)=ax2+bx+c,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:R→Rf:\mathbb R\to\mathbb R
usually written:
f(x)=ax2+bx+c.f(x)=ax^2+bx+c.

Week 2 → coordinate geometry

The graph consists of points:
(x,y)(x,y)
satisfying:
y=ax2+bx+c.y=ax^2+bx+c.
Roots are simply intersections with the xx-axis.
The yy-intercept is the intersection with the yy-axis.
So Week 3 builds directly on Week 2.

59. One deeper mental model

Think of:
y=ax2+bx+cy=ax^2+bx+c
as producing an output yy for every input xx.
The graph is simply a picture of all input-output pairs:
{(x,f(x)):x∈R}.\{(x,f(x)):x\in\mathbb R\}.
That idea applies to every function you'll study later.

60. Formula sheet

Standard form

f(x)=ax2+bx+c\boxed{ f(x)=ax^2+bx+c }

Axis of symmetry

x=−b2a\boxed{ x=-\frac b{2a} }

Vertex

(−b2a,f(−b2a))\boxed{ \left( -\frac b{2a}, f\left(-\frac b{2a}\right) \right) }
or:
(−b2a,−Δ4a)\boxed{ \left( -\frac b{2a}, -\frac{\Delta}{4a} \right) }

Vertex form

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

Quadratic formula

x=−b±b2−4ac2a\boxed{ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} }

Discriminant

Δ=b2−4ac\boxed{ \Delta=b^2-4ac }

Sum of roots

α+β=−ba\boxed{ \alpha+\beta=-\frac ba }

Product of roots

αβ=ca\boxed{ \alpha\beta=\frac ca }

61. What you should actually master

By the end of Week 3, you should be able to:
  1. Recognize a quadratic function.
  2. Identify a,b,ca,b,c correctly.
  3. Determine whether its parabola opens up or down.
  4. Find the yy-intercept.
  5. Find roots by factorization.
  6. Use the quadratic formula.
  7. Calculate and interpret the discriminant.
  8. Find the axis of symmetry.
  9. Find the vertex.
  10. Determine minimum/maximum value.
  11. Find the range.
  12. Convert between standard, factored and vertex forms.
  13. Use root sum/product relations.
  14. Sketch a parabola from algebraic information.
  15. 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 xx-coordinate for y=ax2+bx+cy = ax^2 + bx + c :: x=−b2ax = -\frac{b}{2a}
  • Discriminant formula for ax2+bx+c=0ax^2 + bx + c = 0 :: D=b2−4acD = b^2 - 4ac
  • Roots nature when discriminant D>0D > 0 :: Two distinct real roots
  • Roots nature when discriminant D=0D = 0 :: One real repeated root (tangent to axis)
  • Roots nature when discriminant D<0D < 0 :: Two complex conjugate roots (no x-intercepts)
  • Vieta's formula for sum of roots :: α+β=−ba\alpha + \beta = -\frac{b}{a}
  • Vieta's formula for product of roots :: αβ=ca\alpha \beta = \frac{c}{a}
  • Parabola opening direction if a>0a > 0 :: Upward (has a global minimum at vertex)
  • Parabola opening direction if a<0a < 0 :: Downward (has a global maximum at vertex)

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.
1

Question 1

What is the xx-coordinate of the vertex of the parabola f(x)=2x2−8x+11f(x) = 2x^2 - 8x + 11?
NAT
3 Unit Assessment
2

Question 2

If α\alpha and β\beta are the roots of 2x2−6x+5=02x^2 - 6x + 5 = 0, what is the value of α+β\alpha + \beta?
MCQ
3 Unit Assessment
A. 3
B. -3
C. 5/2
D. -5/2

Active Recall Session

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