Grade by Grade
A. Characteristics of Functions · Quadratic Functions

Quadratic Functions in Three Forms

A2.1 A2.2About 65 minutes
Learning goals
  • I can evaluate and interpret a quadratic function written in function notation.
  • I can read the zeros, vertex, axis of symmetry and y-intercept from the form that shows them.
  • I can move between standard, vertex and factored form.
  • I can decide whether a quadratic function has a maximum or a minimum value.

Words to know

quadratic function
A function that can be written f(x) = ax² + bx + c with a ≠ 0. Its graph is a parabola.
function notation
Writing f(x) for the output of function f at input x. f(3) is the output when x = 3.
zero of a function
An input x with f(x) = 0. The zeros are the x-intercepts of the graph.
standard form
f(x) = ax² + bx + c.
vertex form
f(x) = a(x − h)² + k, with vertex (h, k).
factored form
f(x) = a(x − r)(x − s), with zeros r and s.
maximum / minimum value
The greatest or least output of a function. For a quadratic it is the y-coordinate of the vertex.

One function, three outfits

Picture a snowboarder launching off the lip of a half-pipe at a hill near Collingwood. Her height above the lip rises, levels off and falls. A quadratic function describes that path, and the same function can be written in three different ways. Each way is like an outfit for a different occasion: none changes who the function is, but each one shows off a different feature.

In this unit you will use those forms to find maximum and minimum values, decide how many zeros a function has, find equations from zeros and a point, and find where a line meets a parabola. This first lesson connects what you learned in Grade 10 to the function notation of Grade 11.

Remember from Grade 10

In MPM2D you graphed parabolas, found the vertex, axis of symmetry and zeros, expanded and factored quadratic expressions, completed the square (with whole-number halves), used the quadratic formula and the discriminant, and solved problems about thrown objects, area and revenue. There you wrote relations as y = …. In Grade 11 you write them as functions: f(x) = …. The sibling unit Functions and Equivalent Expressions explains function notation, domain and range in detail.

Function notation for quadratics

Writing f(x) = 2x² − 8x + 6 names the function f and tells you how to get an output from an input x.

  • f(3) means "the output when x = 3": f(3) = 2(9) − 8(3) + 6 = 18 − 24 + 6 = 0.
  • f(x) = 0 asks "which inputs give an output of 0?" Those inputs are the zeros (the x-intercepts).
  • f(0) is the y-intercept. Here f(0) = 6.
  • The maximum or minimum value of f is the output at the vertex: if the vertex is at x = h, the value is f(h).

Be careful: f(3) is not f times 3. The brackets hold an input.

One function, three equivalent formsStandard formf(x) = ax² + bx + cVertex formf(x) = a(x − h)² + kFactored formf(x) = a(x − r)(x − s)
The three forms of a quadratic function. Every quadratic can be written in standard and vertex form; it has a factored form with real numbers only when it has real zeros.
FormWhat you can read directlyExample
Standard: f(x) = ax² + bx + cy-intercept c; direction of opening (sign of a)f(x) = 2x² − 8x + 6
Vertex: f(x) = a(x − h)² + kvertex (h, k); axis x = h; max or min value kf(x) = 2(x − 2)² − 2
Factored: f(x) = a(x − r)(x − s)zeros r and s; axis x = (r + s)/2f(x) = 2(x − 1)(x − 3)
These three equations describe the same function. Check: each one gives f(0) = 6.

The value of a

The number a is the same in all three forms. It tells you:

  • a > 0: the parabola opens up, so the vertex is a minimum.
  • a < 0: the parabola opens down, so the vertex is a maximum.
  • The size of a tells how stretched the graph is: |a| > 1 is narrower than y = x², and 0 < |a| < 1 is wider.
Worked example

Example 1: From factored form to everything

For f(x) = 2(x − 1)(x − 3), find the zeros, the axis of symmetry, the vertex, the y-intercept and the standard and vertex forms.

  1. Zeros. f(x) = 0 when x − 1 = 0 or x − 3 = 0, so x = 1 and x = 3.
  2. Axis of symmetry. It is halfway between the zeros: x = (1 + 3) ÷ 2 = 2.
  3. Vertex. f(2) = 2(2 − 1)(2 − 3) = 2(1)(−1) = −2, so the vertex is (2, −2). Since a = 2 > 0, −2 is the minimum value.
  4. Standard form. Expand: 2(x² − 3x − x + 3) = 2(x² − 4x + 3) = 2x² − 8x + 6. The y-intercept is 6.
  5. Vertex form. Use a = 2 and the vertex (2, −2): f(x) = 2(x − 2)² − 2.
  6. Check vertex form: 2(x² − 4x + 4) − 2 = 2x² − 8x + 8 − 2 = 2x² − 8x + 6 ✓.

Zeros 1 and 3; axis x = 2; minimum value −2 at (2, −2); f(0) = 6; f(x) = 2x² − 8x + 6 = 2(x − 2)² − 2.

−1012345−3−2−1012345678xy(1, 0)(3, 0)vertex (2, −2)y-intercept (0, 6)axis x = 2
The graph of f(x) = 2(x − 1)(x − 3). The axis of symmetry passes halfway between the zeros, and the minimum value −2 sits on it.
Worked example

Example 2: From vertex form to the zeros

For g(x) = −(x + 2)² + 9, state the maximum or minimum value, find the zeros and write the function in factored form.

  1. Read the vertex: x + 2 = x − (−2), so h = −2 and k = 9. The vertex is (−2, 9).
  2. a = −1 < 0, so the parabola opens down and 9 is the maximum value (at x = −2).
  3. Zeros: solve g(x) = 0. −(x + 2)² + 9 = 0, so (x + 2)² = 9.
  4. Take square roots (both signs!): x + 2 = 3 or x + 2 = −3, so x = 1 or x = −5.
  5. Factored form: g(x) = −(x − 1)(x + 5). Check: −(x² + 4x − 5) = −x² − 4x + 5, and −(x² + 4x + 4) + 9 = −x² − 4x + 5 ✓.
  6. The y-intercept is g(0) = 5.

Maximum value 9 at x = −2; zeros −5 and 1; g(x) = −(x − 1)(x + 5).

−6−5−4−3−2−1012−2−101234567891011xy(−5, 0)(1, 0)max (−2, 9)(0, 5)
The graph of g(x) = −(x + 2)² + 9 opens down. Its maximum value is 9, and its zeros −5 and 1 are each 3 units from the axis x = −2.
Worked example

Example 3: What does the notation mean in context?

A ball is tossed straight up. Its height, in metres, after t seconds is h(t) = −5t² + 20t + 1. Find and interpret h(0), h(2) and the solutions of h(t) = 16.

  1. h(0) = 1: the ball starts 1 m above the ground (the height of the thrower's hand in this model).
  2. h(2) = −5(4) + 20(2) + 1 = −20 + 40 + 1 = 21: after 2 s the ball is 21 m high.
  3. h(t) = 16 means −5t² + 20t + 1 = 16, so −5t² + 20t − 15 = 0.
  4. Divide by −5: t² − 4t + 3 = 0, so (t − 1)(t − 3) = 0 and t = 1 or t = 3.
  5. The ball is 16 m high twice: at 1 s on the way up and at 3 s on the way down. Halfway between is t = 2, so the maximum height is h(2) = 21 m.

h(0) = 1 m (start), h(2) = 21 m (the maximum), and h(t) = 16 at t = 1 s and t = 3 s.

Which form should I use?

Need the zeros? Use factored form. Need the maximum or minimum? Use vertex form, or find the axis first. Need the y-intercept or to compare with data? Standard form is quickest. Most of this unit is about moving between forms quickly and accurately.

Hands-on activity

Three-forms matching game

You need: 12 index cards (or squares cut from paper; take care with scissors) and a pencil.

  1. On separate cards write these four functions in standard form: x² − 8x + 12, x² + 2x − 8, 2x² − 8x + 6, −x² − 4x + 5.
  2. On four more cards write their vertex forms: (x − 4)² − 4, (x + 1)² − 9, 2(x − 2)² − 2, −(x + 2)² + 9.
  3. On the last four cards write their factored forms: (x − 2)(x − 6), (x + 4)(x − 2), 2(x − 1)(x − 3), −(x − 1)(x + 5).
  4. Shuffle all 12 cards. Match each standard form with its vertex and factored forms by expanding on scrap paper. Time yourself, then challenge a family member.

What to look for: x² − 8x + 12 = (x − 4)² − 4 = (x − 2)(x − 6); x² + 2x − 8 = (x + 1)² − 9 = (x + 4)(x − 2); 2x² − 8x + 6 = 2(x − 2)² − 2 = 2(x − 1)(x − 3); −x² − 4x + 5 = −(x + 2)² + 9 = −(x − 1)(x + 5). A fast shortcut: matching cards must have the same value at x = 0.

Just for fun

Why did the quadratic function bring three outfits on the school trip?

It wanted to look its best in standard, vertex and factored form.

Your turn

1

For f(x) = x² − 5x + 4, find f(0), f(1) and f(−2).

2

For f(x) = 3(x + 1)(x − 5), state the zeros and the axis of symmetry.

3

Find the minimum value of f(x) = 3(x + 1)(x − 5).

4

Write f(x) = 3(x + 1)(x − 5) in standard form and in vertex form.

5

For g(x) = −2(x − 3)² + 8, state the vertex, whether it is a maximum or minimum, and the y-intercept.

6

Find the zeros of g(x) = −2(x − 3)² + 8.

7

h(t) = −5t² + 20t + 1 gives a ball's height in metres after t seconds. Find h(1) and explain what it means.

More practice: warm-up

1

For f(x) = x² + 4, find f(3).

2

Is the vertex of f(x) = −4x² + x + 1 a maximum or a minimum?

3

State the vertex of f(x) = (x − 6)² + 1.

4

State the zeros of f(x) = (x − 7)(x + 2).

5

What is the y-intercept of f(x) = 5x² − 3x − 9?

More practice: on your own

1

Write f(x) = (x + 4)(x − 2) in vertex form.

2

Find the zeros of f(x) = (x − 5)² − 16 and write it in factored form.

3

f(x) = 0.5(x − 2)(x − 8). Find the minimum value and write f in standard form.

4

For f(x) = −(x − 4)(x + 2), find f(1), and explain why f(1) is the maximum value.

5

For h(t) = −5t² + 20t + 1, solve h(t) = 1 and interpret the answers.

Challenge

1

A quadratic function has zeros −3 and 7 and a maximum value of 50. Write it in factored form and vertex form.

2

Explain why f(x) = (x − 1)² + 4 has no factored form using real numbers.

3

f(x) = ax² + bx + c has f(0) = 6, f(1) = 0 and f(3) = 0. Find a, b and c.

Check your understanding

Quiz

1.

For f(x) = x² − 3x + 2, what is f(−1)?

2.

Which form shows the zeros directly?

3.

What is the vertex of f(x) = −3(x + 4)² − 1?

4.

f(x) = 2(x − 2)(x − 6). What is the minimum value of f?

5.

h(t) is the height in metres of a ball after t seconds. What does h(3) = 12 mean?

6.

Which function has a maximum value?

For parents and teachers

Big idea: This lesson reconnects Grade 10 quadratic skills to Grade 11 function notation. It supports A2.1 (zeros) and A2.2 (maximum and minimum values).

How to help: Ask your teen to "translate" notation into words: "What does f(2) = 21 mean?" Ask which form they would choose for a question and why.

Common difficulties:

  • Reading the vertex of a(x − h)² + k with the wrong sign for h: (x + 2)² means h = −2.
  • Forgetting the ± when taking a square root, which loses one zero.
  • Treating f(3) as multiplication.
  • Forgetting a when converting factored form to vertex form (the a must stay in front).