Families of Quadratic Functions

Knowing where a parabola crosses the $x$-axis isn’t quite enough to pin down its equation: a whole family of parabolas share those zeros. One extra point picks out the exact member you want. This is how you find the equation of an arch, a bridge, or a ball’s path from a few measurements.

Key ideas

A family with the same zeros

In factored form, $y = a(x - r)(x - s)$ has zeros $r$ and $s$ for every value of $a$ (except $0$). Changing $a$ stretches the parabola and can flip it, but the zeros stay put.

Every member of $y = a(x + 1)(x - 3)$ crosses the $x$-axis at $-1$ and $3$.

All the members also share the same axis of symmetry, $x = \dfrac{r + s}{2}$, halfway between the zeros.

Finding one member

To find the specific parabola through a known point:

1. Write the family: $y = a(x - r)(x - s)$.
2. Substitute the point’s $x$ and $y$.
3. Solve for $a$.
4. Write the equation (and expand to standard form if asked).

Zeros that aren’t whole numbers

If the zeros are $p + \sqrt{q}$ and $p - \sqrt{q}$, multiply the factors using the difference of squares:

$$\big(x - (p + \sqrt{q})\big)\big(x - (p - \sqrt{q})\big) = (x - p)^2 - q$$

Worked examples

Example 1: Writing a family

Write the family of quadratics with zeros $2$ and $-5$.

Solution. Each zero gives a factor that equals $0$ at that zero: $x - 2$ and $x + 5$.

$$y = a(x - 2)(x + 5), \qquad a \ne 0$$

Example 2: Finding a member through a point

Find the quadratic with zeros $3$ and $-1$ that passes through $(1, 12)$. Write it in standard form.

Solution. The family is $y = a(x - 3)(x + 1)$. Substitute $x = 1$, $y = 12$:

$$12 = a(1 - 3)(1 + 1) = a(-2)(2) = -4a \quad\Rightarrow\quad a = -3$$ $$y = -3(x - 3)(x + 1) = -3(x^2 - 2x - 3) = -3x^2 + 6x + 9$$

Example 3: Irrational zeros

Find the quadratic with zeros $4 \pm \sqrt{2}$ and $y$-intercept $28$.

Solution. Multiply the factors with the difference of squares:

$$\big(x - (4 + \sqrt{2})\big)\big(x - (4 - \sqrt{2})\big) = (x - 4)^2 - 2 = x^2 - 8x + 14$$

So the family is $y = a(x^2 - 8x + 14)$. At $x = 0$, $y = 14a = 28$, so $a = 2$:

$$y = 2x^2 - 16x + 28$$

Example 4: An arch

A parabolic arch is $40$ m wide at the ground and $10$ m tall at its centre. Find an equation for it.

Solution. Put the left foot of the arch at the origin, so the zeros are $0$ and $40$, and the top is at $(20, 10)$:

$$y = a(x - 0)(x - 40) = ax(x - 40)$$ $$10 = a(20)(20 - 40) = -400a \quad\Rightarrow\quad a = -\frac{1}{40}$$ $$y = -\frac{1}{40}x(x - 40)$$

where $x$ and $y$ are in metres.

Common mistakes

Getting the signs of the factors wrong. A zero at $-5$ gives the factor $(x + 5)$, not $(x - 5)$.

Forgetting $a$. $y = (x - 3)(x + 1)$ is only one member of the family. Include $a$ and solve for it.

Treating the vertex as a zero. In Example 4, the top of the arch $(20, 10)$ is the extra point used to find $a$; the zeros are where the arch meets the ground.

Arithmetic slips when solving for $a$. Substitute carefully, multiply the brackets first, then divide. Check by putting the point back in.

Practice

1. (Warm-up) Write the family of quadratics with zeros $0$ and $6$.

Solution

$y = ax(x - 6)$, $a \ne 0$.

2. (Warm-up) Give the zeros and the axis of symmetry of $y = -2(x + 3)(x - 5)$.

Solution

Zeros $-3$ and $5$. Axis of symmetry $x = \dfrac{-3 + 5}{2} = 1$.

3. (Warm-up) Which of these are in the same family?

• (a) $y = 2(x - 1)(x + 4)$
• (b) $y = -(x + 4)(x - 1)$
• (c) $y = 3(x + 1)(x - 4)$

Solution

(a) and (b): both have zeros $1$ and $-4$. (c) has zeros $-1$ and $4$, so it’s in a different family.

4. (Core) Find the quadratic with zeros $-2$ and $4$ that passes through $(2, -16)$. Write it in standard form.

Solution

$$-16 = a(2 + 2)(2 - 4) = -8a \quad\Rightarrow\quad a = 2$$$$y = 2(x + 2)(x - 4) = 2x^2 - 4x - 16$$

5. (Core) A quadratic has zeros $1$ and $5$ and $y$-intercept $-10$. Find its equation and its maximum value.

Solution

$y = a(x - 1)(x - 5)$. At $x = 0$: $a(-1)(-5) = 5a = -10$, so $a = -2$.

$$y = -2(x - 1)(x - 5)$$

The vertex is at $x = 3$: $y = -2(2)(-2) = 8$. The maximum value is $8$.

6. (Core) Find the quadratic with $x$-intercepts $\tfrac{3}{2}$ and $-1$ that passes through $(0, 6)$. Write it in standard form.

Solution

$y = a\left(x - \tfrac{3}{2}\right)(x + 1)$. At $x = 0$: $a\left(-\tfrac{3}{2}\right)(1) = 6$, so $a = -4$.

$$y = -4\left(x - \tfrac{3}{2}\right)(x + 1) = -4\left(x^2 - \tfrac{1}{2}x - \tfrac{3}{2}\right) = -4x^2 + 2x + 6$$

7. (Core) Find the quadratic with zeros $2 \pm \sqrt{3}$ that passes through $(0, 3)$.

Solution

$$(x - 2)^2 - 3 = x^2 - 4x + 1$$

So $y = a(x^2 - 4x + 1)$. At $x = 0$: $a = 3$.

$$y = 3x^2 - 12x + 3$$

8. (Challenge) A doorway is shaped like a parabola, $2$ m wide at the floor and $3$ m tall in the middle. Can a box $1.2$ m wide and $1.8$ m tall be pushed straight through it (centred)?

Solution

Put the centre of the floor at the origin, so the zeros are $-1$ and $1$, and the top is at $(0, 3)$:

$$y = a(x + 1)(x - 1), \qquad 3 = a(1)(-1) \quad\Rightarrow\quad a = -3$$

The box reaches $0.6$ m on each side of centre. At $x = 0.6$:

$$y = -3(0.6^2 - 1) = -3(-0.64) = 1.92$$

The doorway is $1.92$ m tall there, which is more than $1.8$ m, so yes, the box fits, with $12$ cm to spare at the corners.

9. (Challenge) Explain why every member of $y = a(x - 1)(x - 5)$ has its vertex on the line $x = 3$. Then find the member with vertex $(3, 12)$.

Solution

Every member has zeros $1$ and $5$, and a parabola’s vertex is halfway between its zeros, at $x = 3$. Changing $a$ only changes how high or low the vertex is.

For the vertex $(3, 12)$: $12 = a(2)(-2) = -4a$, so $a = -3$, giving $y = -3(x - 1)(x - 5)$.