Quadratics 1 · The three forms of a quadratic

Module 14 of 46Lesson 1 of 6 · Quadratics 1 · The three forms of a quadratic

Quadratic functions, equations & inequalities · Essential understanding 1

The three forms of a quadratic

IB AA SL 2.6with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: A quadratic function makes a smooth U-shaped curve called a parabola. The same curve can be written in three forms, and each form shows you something different: the y-intercept, the x-intercepts, or the turning point. Choose the form that gives you the answer for free.

By the end of this lesson you can:

recognise $f(x) = ax^2 + bx + c$ , $f(x) = a(x - h)^2 + k$ and $f(x) = a(x - p)(x - q)$
write down the y-intercept, the x-intercepts and the vertex from the right form
find the axis of symmetry $x = -\dfrac{b}{2a}$ and the vertex of any quadratic
sketch a parabola showing its key features, with no calculator
find the equation of a parabola from information given about it

🧑🏾 Imagine this…

Kagiso kicks a football from the ground in Maun. It rises, slows down at the top, then falls back. If you graph height against time you get an upside-down U. The path is symmetric: the way up is a mirror image of the way down. The highest point is the vertex, and the vertical mirror line through it is the axis of symmetry.

Quadratic function

a function whose highest power of

x

x^2

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

with

a \neq 0

Parabola

the U-shaped graph of a quadratic. If

a > 0

it opens upwards (a valley); if

a < 0

it opens downwards (a hill).

Vertex

the turning point of the parabola: the minimum point of a valley or the maximum point of a hill.

Axis of symmetry

the vertical line through the vertex. The parabola is a mirror image on either side of it.

Roots / zeros / x-intercepts

the values of

x

where

f(x) = 0

, that is, where the curve crosses the

x

-axis. A quadratic has 0, 1 or 2 of them.

★ The three forms

Form	Looks like	You can read off
Standard	$f(x) = ax^2 + bx + c$	y-intercept $(0, c)$ ; axis $x = -\dfrac{b}{2a}$
Vertex (turning-point)	$f(x) = a(x - h)^2 + k$	vertex $(h, k)$ ; axis $x = h$
Intercept (factorised)	$f(x) = a(x - p)(x - q)$	x-intercepts $p$ and $q$ ; axis $x = \dfrac{p + q}{2}$

In every form, $a$ tells the shape: $a > 0$ opens up, $a < 0$ opens down, and a bigger $|a|$ makes a narrower curve.

★ Finding the vertex from standard form

The axis of symmetry is $x = -\dfrac{b}{2a}$ .
The vertex is $\left(-\dfrac{b}{2a},\; f\!\left(-\dfrac{b}{2a}\right)\right)$ : find the $x$ -coordinate first, then put it back into $f$ to get the $y$ -coordinate.

⚠️ Watch out

In vertex form the bracket is $(x - h)$ , so $f(x) = (x + 3)^2 - 5$ has vertex $(-3, -5)$ , not $(3, -5)$ . The sign inside the bracket is the opposite of the $x$ -coordinate.
In intercept form, $f(x) = (x + 2)(x - 6)$ has roots $x = -2$ and $x = 6$ . Set each bracket to zero to find them.
The number $a$ is the same in all three forms. When you rewrite a quadratic, $a$ never changes.

💡 How to sketch a parabola (Paper 1)

A sketch needs shape (up or down from the sign of a), the y-intercept, the x-intercepts (if any) and the vertex, all labelled with their coordinates. It does not have to be to scale, but the turning point must be in the right place relative to the intercepts.

🧪 Try it yourselfMove the parabola with a, b and c

Start with f(x) = x² − 4x + 3. Slide c and watch only the y-intercept move. Slide a to flip and squeeze the curve. Slide b and watch the vertex travel along. Check: the vertex is always at x = −

\dfrac{b}{2a}

. For a = 1, b = −4, c = 3 the vertex is (2, −1) and the roots are 1 and 3.

Worked examples

Example 1.1Reading features from standard form

For

f(x) = 2x^2 - 8x + 3

, find (a) the

y

-intercept, (b) the equation of the axis of symmetry, (c) the coordinates of the vertex. Say whether the vertex is a maximum or a minimum.

🧠 Think it through

Here $a = 2$ , $b = -8$ , $c = 3$ .
The y-intercept is where $x = 0$ , which is just $c$ .
Axis: $x = -\dfrac{b}{2a}$ . Then substitute that $x$ into $f$ for the vertex's y-value.
$a = 2 > 0$ , so the parabola opens upwards and the vertex is a minimum.

✍️ By hand

(a)

f(0) = 3

, so the y-intercept is

(0, 3)

(b)

x

-\dfrac{-8}{2(2)}

\dfrac{8}{4}

2

(c)

f(2)

2(4) - 8(2) + 3

8 - 16 + 3

-5

. Vertex

(2, -5)

✅ Answer (a)

(0, 3)

(b)

x = 2

(2, -5)

, a minimum (because

a > 0

Example 1.2Reading features from intercept form

A parabola has equation

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

. Find the

x

-intercepts, the

y

-intercept, the axis of symmetry and the vertex, then sketch it.

🧠 Think it through

Set each bracket to zero for the x-intercepts.
The axis of symmetry is exactly halfway between the two roots.
Substitute the axis value into the equation to get the vertex height. Substitute $x = 0$ for the y-intercept.
The coefficient of $x^2$ is $-1$ (from the minus sign out front), so the curve opens downwards.

✍️ By hand

x-intercepts:

x + 1 = 0

x - 5 = 0

, so

x = -1

and

x = 5

Axis:

x = \dfrac{-1 + 5}{2} = 2

Vertex:

y

-(2 + 1)(2 - 5)

-(3)(-3)

9

, so

(2, 9)

, a maximum.

y-intercept:

y = -(1)(-5) = 5

, so

(0, 5)

✅ Answer x-intercepts

(-1, 0)

and

(5, 0)

; y-intercept

(0, 5)

; axis

x = 2

; vertex

(2, 9)

. The sketch is an upside-down U through those four points.

Check: expanding gives y = −x² + 4x + 5, and

\dfrac{-4}{2 \times (-1)} = 2

, the same axis. ✓

Example 1.3Reading features from vertex form

The function

f(x) = -2(x + 1)^2 + 8

is given in vertex form. (a) Write down the vertex and the axis of symmetry. (b) Find the

y

-intercept. (c) Find the

x

-intercepts. (d) Expand

f(x)

into the form

ax^2 + bx + c

🧠 Think it through

Compare with $a(x - h)^2 + k$ : the bracket is $(x - (-1))$ , so $h = -1$ and $k = 8$ .
For (c) put $f(x) = 0$ and solve by isolating the squared bracket.
For (d) expand $(x + 1)^2 = x^2 + 2x + 1$ first, then multiply by $-2$ .

✍️ By hand
(a) Vertex , axis . Since , it is a maximum.
(b) . So the y-intercept is .
(c)  =  =  = , so  or .
(d)  =  = .
Check: axis  ✓ and  matches (b) ✓.

✅ Answer (a)

(-1, 8)

x = -1

(b)

(0, 6)

(c)

x = 1

x = -3

(d)

f(x) = -2x^2 - 4x + 6

Remember

\pm

: taking a square root gives two answers,

+2

and

-2

Example 1.4Finding the equation from information

(a) A parabola crosses the

x

-axis at

x = -2

and

x = 6

and passes through

(0, -6)

. Find its equation in the form

f(x) = a(x - p)(x - q)

, and find its vertex. (b) Another parabola has vertex

(3, -4)

and passes through

(5, 4)

. Find its equation in vertex form and in standard form.

🧠 Think it through

(a) Use intercept form with $p = -2$ , $q = 6$ , leaving $a$ unknown. Substitute the extra point to find $a$ .
(b) Use vertex form with $h = 3$ , $k = -4$ , leaving $a$ unknown. Substitute $(5, 4)$ .
Use the form that matches the information you were given.

✍️ By hand

(a)

f(x) = a(x + 2)(x - 6)

. At

(0, -6)

-6 = a(2)(-6) = -12a

, so

a = \dfrac{1}{2}

f(x) = \dfrac{1}{2}(x + 2)(x - 6)

. Axis

x = \dfrac{-2 + 6}{2} = 2

;

f(2)

\dfrac{1}{2}(4)(-4)

-8

. Vertex

(2, -8)

(b)

f(x) = a(x - 3)^2 - 4

. At

(5, 4)

4 = a(2)^2 - 4 = 4a - 4

, so

a = 2

Vertex form:

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

. Expand:

2(x^2 - 6x + 9) - 4 = 2x^2 - 12x + 14

✅ Answer (a)

f(x) = \dfrac{1}{2}(x + 2)(x - 6)

, vertex

(2, -8)

. (b)

f(x)

2(x - 3)^2 - 4

2x^2 - 12x + 14

Example 1.5A kicked football (using the GDC)

Kagiso kicks a football. Its height in metres after

t

seconds is

h(t) = -5t^2 + 12t + 1.4

. Find (a) the greatest height and when it occurs, (b) the time when the ball hits the ground, giving your answer to 3 s.f.

🧠 Think it through

The graph of $h$ is a hill, so the greatest height is at the vertex: $t = -\dfrac{b}{2a}$ .
The ball hits the ground when h = 0, the positive root.
On Paper 2 you can find both with the graph page. The vertex needs no GDC at all.

✍️ By hand

(a)

t

-\dfrac{12}{2(-5)}

1.2

h(1.2)

-5(1.44) + 14.4 + 1.4

8.6

(b) Solve

-5t^2 + 12t + 1.4 = 0

, that is

t = \dfrac{-12 \pm \sqrt{144 + 28}}{-10}

. The positive root is

t = \dfrac{12 + \sqrt{172}}{10} \approx 2.51

🔢 On your TI-Nspire CX II

Open a Graphs page. Type -5x^2+12x+1.4 into f1(x) and set the window to about x: −1 to 3, y: −2 to 10.
menu💻 click menu on keypad▸6: Analyze Graph▸3: Maximum gives the top of the hill. menu💻 click menu on keypad▸6: Analyze Graph▸1: Zero (click to the right of the positive crossing) gives the landing time.

✅ Answer (a) 8.6 m after 1.2 s. (b) t = 2.51 s (3 s.f.).

The other root t = −0.111 is a time before the kick, so it is rejected in context.

Challenge questions

⭐ Challenge 1.6Find b and c from the vertexPaper 1 style

The quadratic

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

has its vertex at

(3, -2)

. Find the values of

b

and

c

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

The axis is $x = -\dfrac{b}{2}$ (since $a = 1$ ), and it must equal 3.
Then use f(3) = −2 to find c.

✍️ By hand

-\dfrac{b}{2} = 3

, so

b = -6

f(3) = 9 - 18 + c = -2

, so

c = 7

Check by vertex form:

(x - 3)^2 - 2 = x^2 - 6x + 7

✓.

✅ Answer

b = -6

c = 7

⭐ Challenge 1.7Profit from selling cooked mealsPaper 2 style

A food stall in Gaborone sells cooked meals at

x

pula each. Its weekly profit, in thousands of pula, is

P(x) = -2(x - 3)(x - 9)

for

0 \le x \le 12

(a) Write down the prices at which the stall breaks even (profit is zero).
(b) Find the price that gives the greatest profit, and the greatest profit.
(c) Find $P(0)$ and explain what it means.
(d) Expand P(x) into the form ax² + bx + c and verify the axis of symmetry from the coefficients.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Intercept form shows the roots immediately.
The vertex is halfway between the roots.
Use the graph page afterwards to check the maximum.

✍️ By hand

(a)

P = 0

when

x = 3

x = 9

: P3 and P9.

(b) Axis

x = \dfrac{3 + 9}{2} = 6

P(6) = -2(3)(-3) = 18

. So P6 gives the greatest profit of 18 thousand pula, that is P18 000.

(c)

P(0) = -2(-3)(-9) = -54

. If the meals were given away free, the stall would lose P54 000 a week.

(d)

P(x)

-2(x^2 - 12x + 27)

-2x^2 + 24x - 54

. Axis

x = -\dfrac{24}{2(-2)} = 6

✓.

🔢 On your TI-Nspire CX II

Graph -2(x-3)(x-9) and use Analyze Graph to check the maximum and zeros.

✅ Answer (a) P3 and P9. (b) Price P6, greatest profit P18 000. (c) P(0) = −54: a loss of P54 000. (d) P(x) = −2x² + 24x − 54, axis x = 6.

Next: Lesson 2 · Completing the square

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Quadratics 1 · The three forms of a quadratic

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

The three forms of a quadratic

Worked examples

Challenge questions