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⚙️ Function Kingdom · Functions

Quadratic Functions & Parabolas

Read a parabola from its equation: find the vertex, the axis of symmetry and the zeros of y = ax2 + bx + c, convert to vertex form by completing the square, and use the vertex to answer maximum and minimum questions.

In short

  • Every parabola is symmetric about the vertical line x = -b/(2a), and the vertex sits on that line.
  • Finding a vertex is two steps: -b/(2a) gives the x-coordinate, and substituting it back into the function gives the y-coordinate.
  • The sign of a decides everything about direction: positive opens upwards with a minimum at the vertex, negative opens downwards with a maximum.
  • Standard form shows the y-intercept, factored form shows the zeros, and vertex form y = a(x - h)2 + k shows the vertex (h, k).
  • Completing the square rewrites a quadratic without changing it, so whatever the new bracket adds must be subtracted again.
  • Any question about a greatest height, largest area or best price is asking for the y-coordinate of a vertex.

One curve, folded down the middle

A quadratic function is any rule of the form y = ax2 + bx + c, with a not equal to 0. Its graph is a parabola, and every parabola has the same shape story: it comes in from one side, turns once, and goes back out the other side.

The turning point is the vertex, and the vertical line through it is the axis of symmetry. If you folded the grid along that line, the two halves of the curve would land exactly on top of each other. That single fact is what makes quadratics easy to describe: you only ever need to know where the fold is and how the curve behaves on one side of it.

The sign of a decides which way the curve turns.

  • a is positive: the parabola opens upwards, like a valley, and the vertex is its lowest point.
  • a is negative: the parabola opens downwards, like a hill, and the vertex is its highest point.

The size of a decides how sharply it turns. y = 3x2 is narrower than y = x2, and y = 0.5x2 is wider. And c is the y-intercept — the height of the curve at x = 0 — which you can read off without any work at all.

Where the fold is: x = -b/(2a)

The axis of symmetry of y = ax2 + bx + c is always the line

x = -b/(2a)

Two details in that formula cause almost every wrong answer, so say them out loud. The minus in front of b belongs to the formula, not to the sign of b. And the denominator is 2a, not a.

For y = x2 - 6x + 5: a = 1 and b = -6, so x = -(-6)/(2 x 1) = 6/2 = 3. The axis of symmetry is the line x = 3.

Finding the vertex is then two steps, not one.

  • Step 1. x = -b/(2a) gives the x-coordinate of the vertex.
  • Step 2. Substitute that x back into the function to get the y-coordinate. It is never read off the equation.

Continuing the example: y = 32 - 6 x 3 + 5 = 9 - 18 + 5 = -4, so the vertex is (3, -4). Written as a pair, x first.

Why does the formula work? A parabola is symmetric, so the vertex sits exactly halfway between any two points at the same height — including the two x-intercepts. Averaging those two intercepts is where -b/(2a) comes from, and it is also a shortcut worth knowing: if a parabola crosses the x-axis at 1 and 7, its axis of symmetry is x = 4.

Zeros: where the curve meets the x-axis

The zeros of a function are the x-values that make y = 0, so on the graph they are exactly the points where the curve crosses the x-axis. Finding them means solving the equation ax2 + bx + c = 0 — which is a job you already know how to do.

If the function is written in factored form the zeros are almost free. For y = (x - 1)(x - 7), a product is 0 only when one of its factors is 0, so x - 1 = 0 or x - 7 = 0, giving x = 1 and x = 7. Notice that each bracket has to be *solved*: (x + 3) gives x = -3, not x = 3.

From standard form you factor first. y = x2 - 6x + 5 needs two numbers with product 5 and sum -6, which are -1 and -5, so y = (x - 1)(x - 5) and the zeros are 1 and 5.

A parabola can cross the x-axis twice, touch it once, or miss it entirely — a hill that never reaches the ground has no real zeros — but it always has exactly one vertex. That is why the vertex, not the zeros, is the reliable thing to find first.

Three forms, three secrets

The same parabola can be written three ways, and each form hands you one feature for free.

  • Standard form, y = ax2 + bx + c: shows the y-intercept c at a glance.
  • Factored form, y = a(x - r1)(x - r2): shows the zeros r1 and r2 at a glance.
  • Vertex form, y = a(x - h)2 + k: shows the vertex (h, k) at a glance.

Read the vertex form carefully: the bracket holds x - h, so y = (x - 4)2 + 1 has its vertex at x = 4, and y = (x + 4)2 + 1 has its vertex at x = -4. The sign inside the bracket is the opposite of the coordinate.

Completing the square is how you move from standard form to vertex form. It is a rewriting, not a solving — nothing is being found, and the function is the same function before and after.

y = x2 - 6x + 5

Halve the coefficient of x: half of -6 is -3, and (x - 3)2 = x2 - 6x + 9. Building that bracket quietly added 9, so 9 must come straight back off:

y = (x - 3)2 - 9 + 5 = (x - 3)2 - 4

Vertex (3, -4), which agrees with the -b/(2a) calculation above. When a is not 1, factor it out of the x terms first: y = 2x2 - 12x + 5 becomes y = 2(x2 - 6x) + 5 before you halve anything.

Greatest and least: what the vertex is for

Because the vertex is the highest or lowest point of the whole curve, any question asking for a greatest or least value is a vertex question.

A flare fired upwards has height h(t) = -5t2 + 20t + 1 metres after t seconds. The t2 coefficient is negative, so this parabola opens downwards and the vertex is the peak. The axis of symmetry is t = -20/(2 x -5) = -20/-10 = 2, so the flare is highest after 2 seconds, at a height of h(2) = -20 + 40 + 1 = 21 metres.

Those are two different answers to two different questions, and mixing them up is the most common slip in the topic. The x-coordinate of the vertex says when or where; the y-coordinate says how much.

The same machinery answers fencing questions. With 40 m of fence and a wall along one side, a paddock of depth w has area A(w) = w(40 - 2w) = -2w2 + 40w. The vertex is at w = -40/(2 x -2) = 10, so the greatest area is A(10) = 10 x 20 = 200 square metres.

From here the discriminant b2 - 4ac tells you how many zeros a parabola has before you go looking for them, and the vertex form is the doorway into transformations, where every parabola is just y = x2 slid, stretched and flipped.

Worked examples

Example 1

Find the vertex and the axis of symmetry of f(x) = 2x2 + 12x + 13.

  1. Read off the coefficients: a = 2, b = 12, c = 13.
  2. Axis of symmetry: x = -b/(2a) = -12 / (2 x 2) = -12/4 = -3, so the axis is the line x = -3.
  3. The y-coordinate is not in the equation — substitute: f(-3) = 2 x (-3)2 + 12 x (-3) + 13.
  4. f(-3) = 2 x 9 - 36 + 13 = 18 - 36 + 13 = -5.
  5. Vertex: (-3, -5). Since a = 2 is positive the parabola opens upwards, so -5 is the least value the function takes.

Example 2

Write y = x2 + 8x + 11 in vertex form by completing the square, and state the vertex.

  1. Look at x2 + 8x on its own and halve the coefficient of x: half of 8 is 4.
  2. Check what that bracket really is: (x + 4)2 = x2 + 8x + 16, so building it added 16 that was never there.
  3. Take the 16 back off: y = (x + 4)2 - 16 + 11.
  4. Tidy the constants: y = (x + 4)2 - 5.
  5. The bracket holds x - h, and x + 4 is x - (-4), so h = -4 and k = -5. The vertex is (-4, -5).

Example 3

A stall sells lanterns. If the price is p gold, the day takings are T(p) = -2p2 + 24p gold. What price gives the greatest takings, and what are they?

  1. The p2 coefficient is -2, which is negative, so the graph is a parabola opening downwards and the vertex is the greatest value.
  2. Axis of symmetry: p = -b/(2a) with a = -2 and b = 24, so p = -24 / (2 x -2) = -24/-4 = 6.
  3. That 6 answers "what price", not "how much" — it is the p-coordinate of the vertex.
  4. Substitute for the takings: T(6) = -2 x 62 + 24 x 6 = -72 + 144 = 72.
  5. A price of 6 gold gives the greatest takings, 72 gold.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Find the vertex of the parabola. y = -x2 - 2x - 1 What are the coordinates of its vertex? Give your answer as the pair (x, y).

Answer: (-1, 0)

  1. a = -1, b = -2, c = -1
  2. Axis of symmetry: x = -b/(2a) = 2 / (-2) = -1
  3. Substitute: y = (-1) x (-1)2 - 2 x (-1) - 1 = 0
  4. Vertex: (-1, 0)

Problem 2

Difficulty 3 of 5

f(x) = 3x2 + 6x + 3 What is the equation of the axis of symmetry of this parabola? Answer in the form x = ...

Answer: x = -1

  1. a = 3, b = 6
  2. x = -b/(2a) = (-6) / 6 = -1
  3. Axis of symmetry: x = -1

Problem 3

Difficulty 4 of 5

f(x) = -x2 - 10x + 11 Find the zeros of this function — the x-values where its graph meets the x-axis. Enter the smaller zero first, as (a, b).

Answer: (-11, 1)

  1. Set f(x) = 0: -x2 - 10x + 11 = 0
  2. Factor: -(x + 11)(x - 1) = 0
  3. (x + 11) = 0 gives x = -11; (x - 1) = 0 gives x = 1.
  4. Zeros: (-11, 1), so the graph meets the x-axis at (-11, 0) and (1, 0).

Common mistakes

  • Dropping the minus sign: computing b/(2a) instead of -b/(2a), so the axis of symmetry comes out on the wrong side of the y-axis.
  • Dividing by a instead of 2a, which lands the axis of symmetry twice as far from the y-axis as it should be.
  • Reading the sign of h straight out of the bracket: saying that y = (x + 4)2 - 5 has its vertex at x = 4, when x + 4 is zero at x = -4.
  • Forgetting to subtract h2 after completing the square, and writing x2 - 6x + 5 as (x - 3)2 + 5 instead of (x - 3)2 - 4.
  • Answering a maximum question with the x-coordinate: saying the flare reaches 2 metres when 2 is the number of seconds, and 21 is the height.
  • Giving one zero instead of two, or copying the numbers out of the brackets: reading (x + 3)(x - 5) as zeros 3 and -5 rather than -3 and 5.
  • Using c as the y-coordinate of the vertex. c is the height of the curve at x = 0, which is the vertex only when the axis of symmetry is x = 0.

What you should be able to do

  • Find the vertex and the axis of symmetry of y = ax2 + bx + c.
  • Find the zeros of a quadratic function and say where its graph crosses the x-axis.
  • Complete the square to write a quadratic in vertex form, and read the vertex from it.
  • Use the vertex to find the greatest height or the largest area in an application.

Where this fits in the curriculum

Common Core

  • HSF-IF.C.7.A

    High school — Graph linear and quadratic functions and show intercepts, maxima and minima.

  • HSF-IF.C.8.A

    High school — Use factoring and completing the square in a quadratic function to show zeros, extreme values and symmetry of the graph, and interpret these in context.

  • HSA-SSE.B.3.B

    High school — Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.

  • HSA-REI.B.4.A

    High school — Use the method of completing the square to transform any quadratic equation in x into the form (x - p)^2 = q.

Ontario

  • MTH1W.C4.1

    Grade 9 de-streamed — Compare the characteristics of graphs, tables of values and equations of linear and non-linear relations.

    MTH1W contrasts non-linear with linear relations but never names the vertex, axis or zeros of a parabola — those are Grade 10 (MPM2D) — so this pairing is a judgement call.

SAT

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