SL 2.6 Connect standard, factored, and vertex forms of a quadratic | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

IB Mathematics: Analysis and Approaches SL — study topic SL 2.6

A quadratic function can be written in different algebraic forms without changing the function or its graph. Each form makes some features easier to read than others. Before working with quadratics, recall that expanding brackets uses the distributive law, and solving an equation means finding values that make its two sides equal. In this lesson, the variable is

x

, and the function’s output is

y

. Unless a context states otherwise, take the domain to be all real values of

x

. The three forms studied here are standard form, factored form, and vertex form.

What you will learn

Recognise standard, factored, and vertex forms of a quadratic and identify the information each form shows most directly.
Convert between forms using expansion, factorisation, and completing the square.
Connect the algebraic forms to a parabola’s intercepts, roots, and vertex.
Use graphing technology to check a conversion while retaining an analytical explanation.

1. Read the three forms

In standard form, a quadratic is written as

y=ax^2+bx+c

, where

a

,

b

, and

c

are constants and

a\ne 0

. The coefficient

a

determines whether the parabola opens upward or downward: it opens upward if

a>0

and downward if

a<0

. The constant

c

is the value of

y

when

x=0

, so the graph meets the vertical axis at

(0,c)

.

Factored form is

y=a(x-r_1)(x-r_2)

. The values

r_1

and

r_2

are roots, also called zeros: setting

y=0

shows that the graph meets the horizontal axis at

x=r_1

and

x=r_2

. The roots can be equal. If they are equal, the graph touches the horizontal axis at that point without crossing it. The same non-zero multiplier

a

controls the opening and affects how steep or broad the graph appears.

Vertex form is

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

. Its vertex is

(h,k)

, the turning point of the parabola. The minus sign inside the bracket matters: the horizontal coordinate is

h

, not

-h

. The value

k

is the vertical coordinate. The graph is symmetric about the vertical line

x=h

.

y=ax^2+bx+c=a(x-r_1)(x-r_2)=a(x-h)^2+k

Standard form shows the vertical-axis intercept directly.
Factored form shows the roots directly.
Vertex form shows the vertex and symmetry line directly.

2. Convert forms and connect their features

To change factored form to standard form, expand the brackets and collect like terms. To change standard form to factored form, find values that make the quadratic equal to zero and write the corresponding factors, where this can be done with real-number factors. Substituting

y=0

is the link between factors and roots.

To change standard form to vertex form, complete the square. First factor

a

from the terms involving

x

, then make the expression in brackets a perfect square by adding and subtracting the square of half the coefficient of

x

. The added and subtracted quantities cancel, so the value of the expression does not change. The resulting squared bracket identifies

h

; the remaining constant identifies

k

.

A useful link from standard form is that the horizontal coordinate of the vertex is

h=-\frac{b}{2a}

. Substituting this value of

x

into the original function gives the vertex’s vertical coordinate. This locates the vertex; completing the square shows directly how the vertex form is obtained.

These forms describe the same graph. For example, the roots in factored form locate where the graph meets the horizontal axis, while the vertex form shows the minimum or maximum point. In a context such as the height of an object over time, the variables may have restricted meanings: time might be non-negative, and height might be measured in metres. Use the context’s domain and units when interpreting the graph.

h=-\frac{b}{2a}

Expansion changes the appearance of an expression, not its values.
Completing the square preserves the function by adding and subtracting the same amount.
A graphing calculator can check a conversion, but the algebra explains why it is correct.

3. Graphs, calculator checks, and exam habits

A graph provides a visual check of algebra. Compare the predicted vertical-axis intercept, roots, and vertex with the corresponding points or features on the graph. If two forms are equivalent, their graphs should coincide over the same domain. A calculator’s displayed coordinates may be rounded, so use exact algebra for exact values and state an appropriate accuracy when reporting approximations.

For a purposeful technology check, enter each form as a separate function and display them on the same axes. Choose a window that includes the roots and vertex; a poor window can hide important features. Then use the calculator’s trace or coordinate tools to inspect points, but confirm their values from the algebra. Technology is useful for checking whether the graph opens upward or downward and whether its key points agree.

In an exam-style response, identify the form you have, state the feature you need, and show a conversion or substitution that supports your answer. Do not infer a root just from a rough graph when an exact factorisation is available. Keep signs visible when reading

(x-h)

, and check by expanding your final form or substituting a known point.

Use algebra to justify features and technology to check the picture.
Read the signs in brackets carefully.
State the domain and units when they are supplied by a context.

Worked example

From standard form to vertex form

Write

y=2x^2-8x+3

in vertex form and state its vertex and opening direction.

Factor the quadratic coefficient
Factor $2$ from the terms containing $x$ . Keep the constant $3$ outside those brackets.
$y=2(x^2-4x)+3$
Complete the square
Half of $-4$ is $-2$ , and its square is $4$ . Add $4$ inside the brackets and subtract its contribution outside; because the bracket is multiplied by $2$ , the outside adjustment is $-2(4)$ .
$y=2(x^2-4x+4)+3-8$
Read the vertex
The bracket is a squared difference. Comparing with vertex form gives $h=2$ and $k=-5$ . Since the squared term has positive coefficient, the parabola opens upward.
$y=2(x-2)^2-5$

Answer: The vertex form is

y=2(x-2)^2-5

. The vertex is

(2,-5)

, and the parabola opens upward.

Check: Expanding gives

2(x^2-4x+4)-5=2x^2-8x+8-5=2x^2-8x+3

, which matches the original.

Worked example

From factored form to standard and vertex forms

For

y=-3(x+1)(x-5)

, find the standard and vertex forms, then state the roots and vertex.

Expand to standard form
Multiply the two brackets first, then multiply every term by $-3$ .
$y=-3(x^2-4x-5)=-3x^2+12x+15$
Complete the square
Factor $-3$ from the terms involving $x$ . Half of $-4$ is $-2$ , whose square is $4$ . Account for the adjustment with the outside coefficient $-3$ .
$y=-3(x^2-4x+4)+15+12$
Identify graph features
Rewrite the perfect square and combine the constants. The factored form gives the roots by setting each factor equal to zero. The negative coefficient means the graph opens downward.
$y=-3(x-2)^2+27$

Answer: The standard form is

y=-3x^2+12x+15

and the vertex form is

y=-3(x-2)^2+27

. The roots are

-1

and

5

, and the vertex is

(2,27)

.

Check: The roots have midpoint

2

, matching the vertex’s horizontal coordinate. Substituting

x=2

into the original gives

-3(3)(-3)=27

.

Worked example

Use a vertex and a point to find the forms

A quadratic has vertex

(3,-4)

and passes through

(1,4)

. Find its vertex, standard, and factored forms.

Use vertex form
The given vertex fixes $h=3$ and $k=-4$ , leaving only the coefficient $a$ to determine.
$y=a(x-3)^2-4$
Substitute the known point
At the point $(1,4)$ , substitute $x=1$ and $y=4$ . Solving the resulting equation gives the value of $a$ .
$4=a(1-3)^2-4\quad\Rightarrow\quad a=2$
Expand and factor
Expand the squared bracket to obtain standard form. To find the roots, set the expression equal to zero and solve; these roots determine the factors.
$y=2(x-3)^2-4=2x^2-12x+14=2(x-1)(x-5)$

Answer: The vertex form is

y=2(x-3)^2-4

, the standard form is

y=2x^2-12x+14

, and the factored form is

y=2(x-1)(x-5)

.

Check: The factored form gives roots

1

and

5

, whose midpoint is

3

. Substitution of

x=1

into the vertex form gives

4

, as required.

Common mistakes and how to avoid them

Reading the vertex’s horizontal coordinate as the sign inside the bracket, so that

(x-3)^2

is said to have coordinate

-3

.

Correction: Compare with

(x-h)^2

; the coordinate is

h

. Thus

(x-3)^2

gives

h=3

.

Forgetting to account for the coefficient outside the brackets when completing the square.

Correction: If the bracket is multiplied by

a

, an adjustment made inside it also has its effect multiplied by

a

. Keep that factor when balancing the expression.

Calling the constant term in standard form a root.

Correction: The constant

c

gives the vertical-axis intercept

(0,c)

. Roots are the

x

-values where

y=0

.

Treating a calculator’s rounded graph coordinates as exact.

Correction: Use algebra for exact values and report calculator-based approximations with a stated accuracy.

Lesson summary

Standard form $y=ax^2+bx+c$ shows the vertical-axis intercept.
Factored form $y=a(x-r_1)(x-r_2)$ shows the roots.
Vertex form $y=a(x-h)^2+k$ shows the vertex $(h,k)$ .
Convert forms by expanding, factorising, or completing the square, and check the result by expansion or substitution.

Check your understanding

Question 1

What is the vertex of

y=-(x+2)^2+7

?

$(-2,7)$
$(2,7)$
$(-2,-7)$
$(0,7)$

Show answer and explanation

(-2,7)

Compare with

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

. Since

x+2=x-(-2)

, the vertex is

(-2,7)

.

Question 2

Which statement about

y=4(x-1)(x+3)

is correct?

Its roots are $1$ and $-3$ .
Its vertical-axis intercept is $4$ .
Its vertex is $(1,-3)$ .
It opens downward.

Show answer and explanation

Its roots are

1

and

-3

.

Setting each factor equal to zero gives roots

1

and

-3

. At

x=0

, the function value is

4(0-1)(0+3)=-12

, so the vertical-axis intercept is

(0,-12)

. Its positive leading coefficient means it opens upward.

Question 3

What is the standard form of

y=2(x-4)^2+1

?

$y=2x^2-16x+33$
$y=2x^2-8x+17$
$y=2x^2-16x+1$
$y=2x^2-8x+33$

Show answer and explanation

y=2x^2-16x+33

Expand

(x-4)^2=x^2-8x+16

, multiply by

2

, then add

1

. This gives

2x^2-16x+33

.

Key terms

Quadratic: A function whose highest power of the variable is $2$ , with a non-zero coefficient for that squared term.
Root: An input value for which the function’s output is zero; it corresponds to a point where the graph meets the horizontal axis. With a repeated root, the graph touches the axis there without crossing it.
Vertex: The turning point of a quadratic graph, shown as $(h,k)$ in vertex form.
Complete the square: Rewrite a quadratic expression using a squared bracket and a constant, while preserving its value.

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SL 2.6 · Connect standard, factored, and vertex forms of a quadratic

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — study topic SL 2.6

What you will learn

1. Read the three forms

2. Convert forms and connect their features

3. Graphs, calculator checks, and exam habits

From standard form to vertex form

From factored form to standard and vertex forms

Use a vertex and a point to find the forms

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review