Year 10 Standard · Unit 3 Functions and quadratics · Lesson 3 of 6

One parabola, three ways to write it

A quadratic function can be written in standard, factorized or vertex form. Each one shows a different feature of the same parabola: the y-intercept, the zeros or the turning point.

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 − 2)(x + 6).
    Answerx² + 4x − 12. The x terms are 6x and −2x, and (−2) × 6 = −12.
  2. From Year 10, Lesson 2: factorize x² − 9x + 20.
    Answer(x − 4)(x − 5). The numbers have product 20 and sum −9.
  3. Work out the number halfway between −3 and 9.
    Answer3. Add them and halve: (−3 + 9) ÷ 2 = 3.
Words used in this lesson
parabola
the U-shaped graph of a quadratic function. For example: the path of a thrown ball.
vertex
the turning point of a parabola, its highest or lowest point. For example: the top of an arch.
axis of symmetry
the vertical line through the vertex that mirrors the graph. For example: x = 2 for a parabola with vertex (2, 7).
zero
an x-value where the function equals 0. For example: where a graph meets the x-axis.
y-intercept
the point where the graph meets the y-axis. For example: found by putting x = 0.
concave
curving up like a cup (a > 0) or down like a cap (a < 0). For example: y = −x² is concave down.

The arch and the ground

An arch over a footpath is a parabola. One engineer describes it by the two points where it meets the ground. Another describes it by its highest point. A third gives the equation as it comes out of a calculation. All three describe the same arch, so each form of the equation should be able to give each feature.

Predict first: an arch meets the ground at 2 m and at 10 m from a marker. Where along the ground is its highest point? Then use the figure: one parabola, written in three forms, each showing a different feature.

1
1
−4
(1, −4)turning point (vertex)
−1 and 3zeros
−3y-intercept

y = (x − 1)² − 4 has its vertex at (1, −4) and opens upward. Expanded, it is y = x² − 2x − 3, so it meets the y-axis at −3. It crosses the x-axis at x = −1 and x = 3.

Show me the steps
  1. Keep a = 1 and k = −4 and slide h to 4. The whole parabola moves to the right: the vertex is (4, −4), the zeros are 2 and 6, and the shape does not change.
  2. Now slide k up to 5. The vertex is (4, 5), above the x-axis, and the parabola opens upward, so it never reaches the axis: there are no zeros.
  3. Set a = −1, h = −2 and k = 4. It opens downward with vertex (−2, 4). The zeros are −4 and 0, and the y-intercept is 0, because the graph passes through the origin.
  4. Set a = 0. The equation becomes y = k, a horizontal line, not a parabola, so there is no vertex. Vertex form needs a not equal to 0.

Standard form y = ax² + bx + c shows the y-intercept c. The axis of symmetry is x = −b ÷ (2a), and the vertex lies on it. (Why: a(x − h)² + k expands to ax² − 2ahx + ah² + k, so b = −2ah, and h = −b ÷ (2a).)

Factorized form y = a(x − p)(x − q) shows the zeros p and q. The axis of symmetry is halfway between them, x = (p + q) ÷ 2.

Vertex form y = a(x − h)² + k shows the vertex (h, k). The parabola opens upward (a lowest point) when a > 0 and downward (a highest point) when a < 0.

It is one function, so every form gives the same graph. To find the vertex from standard form, put x = −b ÷ (2a) into the function to get the y-coordinate.

Two common mistakes. The sign inside the bracket is the opposite of the vertex's x-coordinate: y = (x − 4)² + 1 has its vertex at (4, 1), and y = (x + 4)² + 1 has it at (−4, 1). And the zeros of a(x − p)(x − q) are p and q, not −p and −q.

Worked examples

1. y = −2(x − 3)² + 8. Find the vertex, the y-intercept and the zeros.

  1. Vertex form: the vertex is (3, 8). Since a = −2 is negative, it opens downward and 8 is the maximum.
  2. y-intercept: put x = 0: −2(−3)² + 8 = −18 + 8 = −10.
  3. Zeros: −2(x − 3)² + 8 = 0 gives (x − 3)² = 4, so x − 3 = 2 or −2, and x = 5 or x = 1.

2. Write y = 2x² + 8x + 3 in vertex form.

  1. The axis of symmetry is x = −b ÷ (2a) = −8 ÷ 4 = −2.
  2. Put x = −2 in: 2(4) − 16 + 3 = −5. The vertex is (−2, −5).
  3. With a = 2: y = 2(x + 2)² − 5. Check: 2(x² + 4x + 4) − 5 = 2x² + 8x + 3.

Your go. The first step is given. y = x² − 10x + 9 factorizes as (x − 1)(x − 9). Find its vertex.

  1. The zeros are 1 and 9, so the axis of symmetry is halfway between them: x = (1 + 9) ÷ 2 = 5.
  2. Now put x = 5 into the function to find the y-coordinate of the vertex.
Check the answer

y = 25 − 50 + 9 = −16, so the vertex is (5, −16). Using the factorized form: (5 − 1)(5 − 9) = 4 × (−4) = −16.

With a GDC (optional)

A GDC can draw the graph and find its turning point. Enter y = x² + 10x + 7, graph it, and use the minimum tool (under the graph or calculate menu): it reports the vertex, here (−5, −18). Use the zero tool to find where it meets the x-axis.

The GDC often shows a value such as −4.9999999 for −5, 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. y = 3(x + 2)² − 7. What is its vertex?

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. The parabola y = (x − 2)(x + 8) crosses the x-axis at two points. What is the equation of its axis of symmetry, x = ?

3. y = x² − 8x + 5. Find the y-coordinate of its vertex.

Level 2 · Building

4. The path of a water jet in a fountain in Granada is y = −0.5x² + 4x + 1, where x is the distance from the nozzle and y is the height, both in metres. What is the greatest height of the jet, in metres?

5. A parabola has its vertex at (−1, 4), so y = a(x + 1)² + 4. It passes through the point (1, −4). Find a.

Level 3 · Stretch

6. y = 2x² + bx + 5 has its axis of symmetry at x = 3. Find b.

7. A quadratic function has zeros at x = −1 and x = 5 and passes through (0, 10). It has the form y = a(x + 1)(x − 5). What is the y-coordinate of its vertex?

Something to think about

Slide a to 0. What happens to the graph, and what does that tell you about why a quadratic needs a not equal to 0? Then try a = 3 and a = −3 and describe what a does to the shape.

One way to think about it

At a = 0 the x² term disappears and y = k is a straight line, so a quadratic must have a not equal to 0. A larger |a| makes the parabola narrower, because y changes faster as you move away from the vertex; a smaller |a| makes it wider; and the sign of a decides whether the vertex is a lowest or a highest point.

Exit ticket

8. Which of these is the same function as y = x² − 12x + 27?