Year 10 Extended · Unit 3 Functions, quadratics and transformations · Lesson 3 of 7

The vertex is always findable

Standard, factorized and vertex form describe one parabola. Completing the square turns standard form into vertex form, and doing it once in general gives a formula for the vertex that works for every quadratic.

FunctionsGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: expand (x − 5)² + 1.
    Answerx² − 10x + 26. (x − 5)² = x² − 10x + 25, and then add 1.
  2. From this unit, Lesson 2: factorize x² − 2x − 15.
    Answer(x − 5)(x + 3). The numbers have product −15 and sum −2.
  3. Work out the number halfway between −9 and 5.
    Answer−2. Add them and halve: (−9 + 5) ÷ 2 = −2.
Words used in this lesson
vertex form
a(x − h)² + k, which shows the turning point (h, k). For example: y = (x − 11)² + 10.
completing the square
rewriting a quadratic as a squared bracket plus a number. For example: x² + 10x = (x + 5)² − 25.
turning point
the vertex, where a parabola changes from falling to rising. For example: the top of a thrown ball's path.
least value
the lowest y a function takes. For example: k when a > 0.
greatest value
the highest y a function takes. For example: k when a < 0.
equivalent forms
different expressions that give the same function. For example: x² − 49 and (x − 7)(x + 7).

The cable and its lowest point

A cable on a suspension footbridge in Chengdu hangs in a parabola, and its height above the deck is y = x² − 8x + 19 metres at distance x metres from the first tower. An engineer needs the lowest point of the cable, and she needs it exactly, not read from a graph.

Predict first: how far from the first tower is the lowest point, and how high above the deck is it? Then use the figure: it turns standard form into vertex form for any a, b and c.

1
4
1
(x + 2)² − 3vertex form
(−2, −3)vertex
≈ −3.73 and −0.27zeros

y = x² + 4x + 1: h = −b ÷ (2a) = −2 and k = c − b² ÷ (4a) = 1 − 16 ÷ 4 = −3, so y = (x + 2)² − 3. The vertex is (−2, −3) and it opens upward.

Show me the steps
  1. Set a = 1 (third position), b = −6 and c = 5. Then h = 6 ÷ 2 = 3 and k = 5 − 36 ÷ 4 = −4, so y = (x − 3)² − 4. The vertex is (3, −4) and the zeros are 1 and 5.
  2. Change a to 2 (fourth position), keeping b = −6 and c = 5. Now h = 6 ÷ 4 = 1.5 and k = 5 − 36 ÷ 8 = 0.5, so y = 2(x − 1.5)² + 0.5. The vertex is (1.5, 0.5), above the axis, so there are no zeros.
  3. Set a = −1 (second position), b = 4 and c = 1. Then h = −4 ÷ (−2) = 2 and k = 1 − 16 ÷ (−4) = 5, so y = −(x − 2)² + 5. It opens downward and 5 is the greatest value.
  4. Set a = 1, b = 0 and c = −9. Then h = 0 and k = −9: y = x² − 9, the basic parabola moved down 9. The zeros are −3 and 3.

Completing the square rewrites ax² + bx + c as a(x − h)² + k. Take a out of the x terms, add and subtract the square of half the coefficient, and the first part is a perfect square.

The general result. ax² + bx + c = a(x + b ÷ (2a))² + c − b² ÷ (4a). So the vertex is at x = h = −b ÷ (2a), with y = k = c − b² ÷ (4a). That is the axis of symmetry from Year 9, and it is also where the minimum or maximum occurs.

Each form shows one thing. Standard form shows the y-intercept c, factorized form a(x − p)(x − q) shows the zeros, vertex form shows the vertex. They are one function.

The least value (a > 0) or greatest value (a < 0) of a quadratic is k, so a model's best or worst case is read from the vertex.

Take a out before completing the square. 3x² + 18x + 4 = 3(x² + 6x) + 4, and then 3(x + 3)² − 27 + 4 = 3(x + 3)² − 23. The 27 is 3 × 9, because the 3 multiplies the whole bracket; it is not 9. And the sign inside the bracket is the opposite of the x-coordinate.

Worked examples

1. Write y = x² + 6x − 5 in vertex form.

  1. Half of 6 is 3, and 3² = 9: x² + 6x = (x + 3)² − 9.
  2. So y = (x + 3)² − 9 − 5 = (x + 3)² − 14.
  3. The vertex is (−3, −14), and a = 1 > 0, so −14 is the least value.

2. Find the greatest value of y = −3x² + 6x + 1.

  1. Take −3 out of the x terms: −3(x² − 2x) + 1.
  2. x² − 2x = (x − 1)² − 1, so y = −3[(x − 1)² − 1] + 1 = −3(x − 1)² + 3 + 1.
  3. y = −3(x − 1)² + 4, so the greatest value is 4, at x = 1.

3. Prove the vertex formula.

  1. ax² + bx + c = a[x² + (b ÷ a)x] + c.
  2. x² + (b ÷ a)x = (x + b ÷ (2a))² − b² ÷ (4a²), because half of b ÷ a is b ÷ (2a).
  3. So the function is a(x + b ÷ (2a))² − b² ÷ (4a) + c. Its vertex is x = −b ÷ (2a) and y = c − b² ÷ (4a).

Your go. The first step is given. Write y = 2x² − 20x + 41 in vertex form.

  1. Take 2 out of the x terms: y = 2(x² − 10x) + 41.
  2. Now complete the square inside the bracket and collect the numbers.
Check the answer

x² − 10x = (x − 5)² − 25, so y = 2[(x − 5)² − 25] + 41 = 2(x − 5)² − 9. The vertex is (5, −9). By the formula: h = 20 ÷ 4 = 5 and k = 41 − 400 ÷ 8 = −9.

With a GDC (optional)

A GDC can graph a quadratic and find its turning point. Enter y = x² + 6x + 14, graph it, and use the minimum tool (under the graph or calculate menu): it reports the vertex, here (−3, 5). Use the zero tool for where it meets the x-axis; for this one it never does.

A GDC may show −2.9999999 for −3, because it works with decimals. Round it, and give the exact value from your hand working. Everything here can be done by hand.

Check before you go on

1. Write y = 2x² − 12x + 7 in vertex form.

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. y = 3x² − 12x + 20. What is the least value of y?

3. The vertex of y = 2(x − 5)² − 3 is (h, k). Work out h + k.

Level 2 · Building

4. 2x² + 8x + 9 can be written 2(x + p)² + q. Find q.

5. A football in a park in Medellín is kicked straight up. Its height is h = −5t² + 12t + 1 metres after t seconds. What is its greatest height, in metres?

Level 3 · Stretch

6. The least value of y = x² + bx + 20 is 4. Find the positive value of b.

7. The graph of y = x² + bx + c has its vertex at (6, −2). Find c.

Something to think about

Keep b = 0 and c = 0, and look at the four values of a that the first slider offers. The vertex does not move. What does a change, and what would a change if b were not 0? Then explain why the vertex of y = ax² + c is always on the y-axis.

One way to think about it

With b = 0 the axis x = −b ÷ (2a) is x = 0 whatever a is, so the vertex is (0, c) on the y-axis: a changes only how steep the parabola is, and its sign says whether the vertex is a lowest or a highest point. When b is not 0, a also changes the axis, because h = −b ÷ (2a): a larger size of a, ignoring its sign, squeezes the vertex towards the y-axis.

Exit ticket

8. Which of these is NOT the same function as y = 2(x − 1)² − 18?