Grade by Grade
A. Characteristics of Functions · Quadratic Functions

Finding the Equation from Zeros and a Point

A2.4About 70 minutes
Learning goals
  • I can write a quadratic function from its zeros and one other point.
  • I can multiply factors with conjugate radical zeros quickly.
  • I can handle fractional zeros and double roots.
  • I can model an arch or other real shape with a quadratic function.

Words to know

root
A solution of the equation f(x) = 0. The roots of the equation are the zeros of the function.
conjugates
p + √q and p − √q. The factors they give multiply to (x − p)² − q.
double root
A root that appears twice, as in a(x − 3)². The graph touches the x-axis without crossing.

Building a function to order

Architects, game designers and engineers often know where an arch meets the ground and one more point it must pass through, such as its highest point. From that, they need the equation. In the last lesson you found members of a family when the zeros were integers. Now the zeros can be fractions or radicals, like the curriculum's own example: roots 1 + √5 and 1 − √5, passing through (2, 5).

The method

  1. Write the family: f(x) = a(x − r)(x − s).
  2. If the zeros are radicals, multiply the factors to get a trinomial without radicals.
  3. Substitute the given point and solve for a.
  4. Write the final function, in factored or standard form.

The conjugate shortcut

Irrational zeros of a quadratic with rational coefficients come in conjugate pairs, p + √q and p − √q. Their factors multiply neatly, using a difference of squares:

(x − (p + √q))(x − (p − √q)) = ((x − p) − √q)((x − p) + √q) = (x − p)² − q.

Worked example

Example 1: Irrational roots (the curriculum example)

Find the equation of the quadratic function that passes through (2, 5) if the roots of the corresponding quadratic equation are 1 + √5 and 1 − √5.

  1. Here p = 1 and q = 5, so the factors multiply to (x − 1)² − 5 = x² − 2x + 1 − 5 = x² − 2x − 4.
  2. Family: f(x) = a(x² − 2x − 4).
  3. Substitute (2, 5): 5 = a(4 − 4 − 4) = −4a, so a = −5/4.
  4. f(x) = −(5/4)(x² − 2x − 4) = −(5/4)x² + (5/2)x + 5.
  5. Check: f(2) = −5 + 5 + 5 = 5 ✓. And from Lesson 5, the zeros of x² − 2x − 4 are 1 ± √5 ✓.

f(x) = −(5/4)(x² − 2x − 4), or f(x) = −1.25x² + 2.5x + 5.

−2−101234−3−2−1012345678xy1 − √51 + √5(2, 5)
The graph of f(x) = −1.25(x² − 2x − 4). It crosses the x-axis at 1 − √5 ≈ −1.24 and 1 + √5 ≈ 3.24 and passes through (2, 5).
Worked example

Example 2: A fractional zero

Find the quadratic function with zeros −3 and ½ that passes through (1, 8). Write it in standard form.

  1. Family: f(x) = a(x + 3)(x − ½).
  2. Substitute (1, 8): 8 = a(4)(½) = 2a, so a = 4.
  3. f(x) = 4(x + 3)(x − ½). It is neater to move a factor of 2 into the second bracket: f(x) = 2(x + 3)(2x − 1).
  4. Expand: 2(2x² − x + 6x − 3) = 2(2x² + 5x − 3) = 4x² + 10x − 6.
  5. Check: f(1) = 4 + 10 − 6 = 8 ✓.

f(x) = 4x² + 10x − 6.

Watch the value of a

f(x) = 4(x + 3)(x − ½) and f(x) = 2(x + 3)(2x − 1) are the same function. If you write the family as a(x + 3)(2x − 1) from the start, you will get a = 2, not 4. Both are correct; just do not mix the two forms.

Worked example

Example 3: Using the y-intercept

A quadratic function has zeros 2 + √3 and 2 − √3 and a y-intercept of 3. Find its equation.

  1. p = 2 and q = 3: (x − 2)² − 3 = x² − 4x + 1. Family: f(x) = a(x² − 4x + 1).
  2. The y-intercept is f(0) = a(1) = a, so a = 3.
  3. f(x) = 3(x² − 4x + 1) = 3x² − 12x + 3.

f(x) = 3x² − 12x + 3.

Worked example

Example 4: A double zero

A parabola touches the x-axis only at x = 3 and passes through (5, 8). Find its equation.

  1. Touching (not crossing) at x = 3 means a double root: f(x) = a(x − 3)².
  2. Substitute (5, 8): 8 = a(2)² = 4a, so a = 2.

f(x) = 2(x − 3)².

10 m40 m(5, ?)(0, 0)(40, 0)h(x) = a·x(x − 40)top: h(20) = 10so a = −1/40h(5) = 4.375 m(not to scale)
A parabolic arch 40 m wide and 10 m high. Placing one end at the origin makes the zeros 0 and 40; the top (20, 10) gives a = −1/40.
Worked example

Example 5: An arch

A bridge has a parabolic arch that spans 40 m across a river and rises 10 m at its centre. How high is the arch 5 m from one end?

  1. Place the left end at (0, 0) and the right end at (40, 0). Family: h(x) = ax(x − 40).
  2. The top is above the middle: (20, 10). So 10 = a(20)(−20) = −400a, and a = −1/40.
  3. h(x) = −(1/40)x(x − 40).
  4. h(5) = −(1/40)(5)(−35) = 175/40 = 4.375.
  5. (Placing the origin at the centre instead gives zeros ±20 and h(x) = −(1/40)(x² − 400). The answer at x = −15 is the same.)

About 4.4 m high, 5 m from the end.

Hands-on activity

Model a real arch

You need: a strip of thin cardboard (a cereal box side works) about 50 cm long, two heavy books, a ruler and a pencil.

  1. Stand the books 30 cm apart. Bend the strip and wedge its ends against the books so it forms an arch.
  2. Measure the height at the centre, h, in centimetres.
  3. Model the arch: put the zeros at 0 and 30, so f(x) = ax(x − 30), and use (15, h) to find a.
  4. Use your model to predict the height 5 cm from one end. Then measure it.
  5. How close was your prediction?

What to look for: a bent strip is not exactly a parabola, but it is close. For example, if h = 10 cm then a = −10/225 = −2/45, and the prediction at x = 5 is f(5) = (−2/45)(5)(−25) ≈ 5.6 cm. Expect your measurement to be within about a centimetre.

Just for fun

Give a quadratic two zeros and it shrugs: "That narrows it down to infinitely many of us." Give it one more point and it says, "Fine. It's me."

Your turn

1

Find the quadratic function with zeros 4 and −1 that passes through (2, −12). Write it in standard form.

2

Multiply (x − (3 + √2))(x − (3 − √2)).

3

Find the quadratic function with zeros 3 ± √2 that passes through (3, 4).

4

Find the quadratic function with zeros −2/3 and 2 and y-intercept 4.

5

A parabola touches the x-axis only at x = −1 and passes through (1, −8). Find its equation.

6

Find the quadratic function with zeros ±√5 that passes through (1, 8).

More practice: warm-up

1

Write the factors for zeros 6 and −3.

2

Expand (x − √7)(x + √7).

3

Write the quadratic function with zeros 0 and 5 and a = 3.

4

f(x) = a(x − 1)(x − 4) passes through (0, 8). Find a.

5

Expand (x − (1 + √2))(x − (1 − √2)).

More practice: on your own

1

Find the quadratic function with zeros 2 ± √5 and y-intercept −2.

2

Find the quadratic function with zeros −1 ± √3 that passes through (1, 4).

3

Find the quadratic function with zeros ½ and −4 that passes through (−3, 7). Write it in standard form.

4

A parabolic doorway is 4 m wide at the floor and 3 m tall. Can a box 1.5 m wide and 2.5 m tall pass through if it is pushed through the centre?

5

A parabola has x-intercepts −1 and 5 and its vertex has y-coordinate 18. Find its equation and its y-intercept.

Challenge

1

A quadratic function has zeros 3 ± √2 and a maximum value of 4. Find it.

2

The zeros of a quadratic function add to 6 and multiply to 4, and it passes through (1, 2). Find the function and its exact zeros.

3

Explain why the zeros 2 and 6 together with the point (6, 0) do not determine a unique quadratic function.

Check your understanding

Quiz

1.

A quadratic function has zeros 1 + √3 and 1 − √3. Which expression gives its family?

2.

f(x) = a(x + 2)(x − 5) passes through (0, 20). What is a?

3.

A parabola touches the x-axis only at x = 4 and passes through (6, 12). Its equation is:

4.

You know the zeros of a quadratic function. Which extra information can NOT be used to find a?

5.

A quadratic function has zeros ±√6 and y-intercept −12. Its equation is:

6.

For the curriculum example (roots 1 ± √5, through (2, 5)), what is a?

For parents and teachers

Big idea: Expectation A2.4 (second part): determine the equation of a quadratic function given the real roots of the corresponding equation and a point on the function. Example 1 is the curriculum's own sample problem.

How to help: Ask your teen to check the final equation two ways: substitute the given point, and confirm the zeros (by factoring or the quadratic formula).

Common difficulties:

  • Sign errors when writing factors from zeros.
  • Expanding (x − (1 + √5))(x − (1 − √5)) term by term and making radical errors. The (x − p)² − q shortcut avoids this.
  • Changing the form of a factor (x − ½ versus 2x − 1) partway through, which changes a.
  • Forgetting the double root a(x − h)² when the graph only touches the axis.