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Quadratic Relations of the Form y = ax² + bx + c · Quadratic Relations and Their Graphs

Finding the Equation of a Parabola in Vertex Form

2.4 1.1About 70 minutes
Learning goals
  • I can find the equation y = a(x − h)² + k of a parabola from its graph.
  • I can find the equation from the zeros and the maximum or minimum value, or from a table of values.
  • I can choose axes to model a real parabola, such as an arch, and use the equation to answer questions.
  • I can write a vertex-form model for quadratic data.

Words to know

vertex form
y = a(x − h)² + k; the vertex is (h, k).
model
An equation that describes a real situation, such as the shape of an arch or the path of a ball.
origin
The point (0, 0), where the axes cross; in a real problem you choose where to put it.

Working backwards

So far you started with an equation and drew the graph. Now you will do the reverse: look at a parabola (a graph, a table, a photo of a bridge or a set of data) and write its equation. Vertex form makes this surprisingly quick: you only need the vertex and one other point.

The method

  1. Find the vertex (h, k) from the graph, the table (use symmetry), or the zeros (the axis is halfway between them).
  2. Write y = a(x − h)² + k with h and k filled in.
  3. Choose another point (x, y) on the parabola that is not the vertex. Substitute its coordinates.
  4. Solve for a.
  5. Write the final equation and check with a third point if you can.

Check the sign of your a with the picture: a parabola that opens down must have a negative a.

xy−1012345−4−3−2−1012345678vertex (2, −3)(4, 5)
A parabola with vertex (2, −3) that passes through (4, 5). These two points are enough to find its equation.
Worked example

Example 1: From a graph

Find the equation of the parabola in the graph above.

  1. Vertex (2, −3), so y = a(x − 2)² − 3.
  2. Another point: (4, 5). Substitute: 5 = a(4 − 2)² − 3.
  3. 5 = 4a − 3, so 8 = 4a and a = 2.
  4. Equation: y = 2(x − 2)² − 3.
  5. Check with the step pattern: from the vertex, over 1 up 2 gives (3, −1); over 1 more up 6 gives (4, 5). ✓

y = 2(x − 2)² − 3

Worked example

Example 2: From the zeros and the maximum

A parabola has zeros 1 and 7 and a maximum value of 9. Find its equation.

  1. The axis of symmetry is halfway between the zeros: x = (1 + 7) ÷ 2 = 4.
  2. The maximum value is the y-coordinate of the vertex, so the vertex is (4, 9): y = a(x − 4)² + 9.
  3. Use a zero, (1, 0): 0 = a(1 − 4)² + 9 = 9a + 9, so a = −1.
  4. Equation: y = −(x − 4)² + 9. Check the other zero: −(7 − 4)² + 9 = −9 + 9 = 0. ✓ It opens down, which matches having a maximum.

y = −(x − 4)² + 9

10 m40 m(0, 0)(40, 0)vertex (20, 10)not to scale
A parabolic arch 40 m wide at its base and 10 m high. Placing the origin at the left end of the base makes the vertex (20, 10).
Worked example

Example 3: An arch bridge

A bridge has a parabolic arch that is 40 m wide at the water line and 10 m high in the middle. Write an equation for the arch, then find the height of the arch 5 m from one end.

  1. Choose axes: put the origin at the left end of the arch at the water line. Then the ends are (0, 0) and (40, 0), and the vertex is (20, 10).
  2. y = a(x − 20)² + 10. Substitute (0, 0): 0 = a(400) + 10, so a = −10/400 = −0.025.
  3. Equation: y = −0.025(x − 20)² + 10, for 0 ≤ x ≤ 40.
  4. 5 m from the left end, x = 5: y = −0.025(5 − 20)² + 10 = −0.025(225) + 10 = −5.625 + 10 = 4.375 m.

y = −0.025(x − 20)² + 10; the arch is about 4.4 m high 5 m from an end.

You choose the axes

In a real problem, you decide where the origin goes. Different choices give different (but equally correct) equations. If you put the origin at the top of the arch in Example 3, the vertex is (0, 0) and the equation becomes y = −0.025x², with the ends at (−20, −10) and (20, −10). Choose whatever makes the numbers easiest, and say what you chose.

Worked example

Example 4: From a table of values

Find the equation of the quadratic relation: x = 0, 1, 2, 3, 4, 5 and y = 11, 5, 3, 5, 11, 21.

  1. The values are symmetric about x = 2 (5 and 5, 11 and 11), and y = 3 is the smallest, so the vertex is (2, 3): y = a(x − 2)² + 3.
  2. Substitute (0, 11): 11 = a(4) + 3, so 4a = 8 and a = 2.
  3. Equation: y = 2(x − 2)² + 3.
  4. Check with second differences: first differences −6, −2, 2, 6, 10; second differences 4 = 2a. ✓ And at x = 5: 2(9) + 3 = 21. ✓

y = 2(x − 2)² + 3

Worked example

Example 5: A model for real data

A ball tossed upward was measured every 0.5 s: (0, 1.5), (0.5, 5.2), (1.0, 6.5), (1.5, 5.3), (2.0, 1.6). Write a vertex-form model.

  1. The data is almost symmetric about t = 1.0, and 6.5 is the largest value, so use the vertex (1, 6.5): h = a(t − 1)² + 6.5.
  2. Use the first data point (0, 1.5): 1.5 = a(0 − 1)² + 6.5, so a = 1.5 − 6.5 = −5.
  3. Model: h = −5(t − 1)² + 6.5.
  4. Check another point: t = 2: −5(1) + 6.5 = 1.5, close to the measured 1.6. A model from real data is a curve of best fit, so small differences are expected. (Physics says a ≈ −4.9 for an object thrown on Earth, very close to our −5.)

h = −5(t − 1)² + 6.5 (h in metres, t in seconds).

Just for fun

A funny-but-true fact: the main cables of a suspension bridge, which carry a flat, evenly loaded deck, hang very close to a parabola, but a chain hanging under only its own weight makes a different curve, called a catenary. The two look so alike that people mixed them up for a long time: Galileo compared a hanging chain to a parabola, and the true shape of the catenary was only worked out in 1691 (by Leibniz, Huygens and Johann Bernoulli).

Hands-on activity

Find the equation of a real parabola

You need: a photo with a parabola in it (an arch, a water fountain, a ball's path in a sports photo, or a parabola you draw by tossing a ball and photographing it with a slow shutter or burst mode), a printout or a screen, a ruler and graph paper (or Desmos, which lets you add a photo to the graph).

  1. Draw or imagine axes on the photo. Put the origin somewhere sensible (for example at one end of the arch).
  2. Use the ruler to measure the vertex position and one other point, in centimetres on the picture.
  3. Write the vertex-form equation using the method of this lesson.
  4. Check a third point of the curve against your equation.
  5. Optional: in Desmos, add the image (the + menu), type your equation, and adjust a, h and k with sliders until the curve lies on the arch.

What to look for: a correct method (vertex first, one more point, solve for a), a negative a for an arch that opens down, and a third point that is close to the curve. Measurements from a photo are only approximate, so small errors are normal.

Your turn

1

A parabola has vertex (1, 4) and passes through (3, 12). Find its equation.

2

A parabola has vertex (−2, 6) and passes through (0, 2). Find its equation.

3

A parabola has vertex (0, −5) and passes through (2, −3). Find its equation.

4

A parabola has zeros −3 and 5 and a minimum value of −8. Find its equation.

5

Find the equation from the table: x = −1, 0, 1, 2, 3 and y = 6, 0, −2, 0, 6.

6

A parabola with vertex (4, 0) passes through (6, −8). Find its equation.

More practice: warm-up

1

What do you need to know to find a in y = a(x − h)² + k?

2

A parabola has vertex (3, −1). Write its equation with a still unknown.

3

A parabola opens down. Is a positive or negative?

4

Zeros 2 and 8: what is the x-coordinate of the vertex?

5

Solve 10 = 4a + 2 for a.

More practice: on your own

1

A parabola has vertex (−1, −4) and y-intercept −2. Find its equation.

2

A footbridge arch is 12 m wide and 3 m high. With the origin at the left end, find its equation.

3

A ball's path is a parabola. It is kicked from (0, 0), reaches a maximum height of 8 m at a horizontal distance of 10 m. Find its equation and where it lands.

4

The table shows x = 1, 2, 3, 4, 5 and y = −7, −1, 1, −1, −7. Find the equation.

5

The parabola y = a(x + 3)² + k passes through (−3, 4) and (−1, 12). Find a and k.

Challenge

1

A parabola passes through (0, 7), (2, −1) and (4, 7). Find its equation in vertex form.

2

A tunnel entrance is a parabola 8 m wide at the road and 6 m high in the middle. A truck is 2.5 m wide and 4.2 m tall. Can it pass through if it drives exactly in the centre? Show your reasoning.

3

A parabola has its vertex on the line x = 3, passes through (1, 5) and (2, 2). Find its equation.

Check your understanding

Quiz

1.

A parabola has vertex (3, 1) and passes through (4, 4). What is a?

2.

Which is the equation of a parabola with vertex (−1, 2) through (0, 0)?

3.

A parabola has zeros −2 and 6. Its vertex has x-coordinate…

4.

An arch has vertex (15, 9) and passes through (0, 0). What is a?

5.

A table has y = 9, 4, 1, 0, 1, 4, 9 for x = 0 to 6. Its equation is…

6.

To find a in vertex form, the extra point you substitute must…

For parents and teachers

Big idea: Expectation 2.4 asks students to determine the equation, in the form y = a(x − h)² + k, of a given graph of a parabola. The method is always: vertex → one more point → solve for a → check. It also connects back to expectation 1.1 (a curve of best fit for real data).

How to help: Look for parabolas in photos (bridges, fountains, sports) and challenge your teen to set up axes and an equation. Ask: "Where did you put the origin, and why?"

Common difficulties:

  • Substituting the vertex to find a (it gives no information).
  • Sign errors with h when writing (x − h).
  • Forgetting to square before multiplying by a when substituting.
  • Not checking the sign of a against the picture.