Families of Quadratic Polynomial Functions (VCE SSCE Mathematical Methods): Revision Notes

Worked Example: Using general form with two unknowns

A family of parabolas have rules of the form $y = ax^2 + c$. For the parabola in this family that passes through the points $(1, 7)$ and $(2, 10)$, find the values of $a$ and $c$.

Solution:

When $x = 1$, $y = 7$:

When $x = 2$, $y = 10$:

Subtract equation (1) from equation (2):

Substitute $a = 1$ into equation (1):

Answer: The equation is $y = x^2 + 6$

Explanation: We substitute each point's coordinates into the family equation to create two equations with two unknowns. Then we solve these simultaneously.

Worked Example: Using the discriminant

A family of parabolas have rules of the form $y = ax^2 + bx + 2$, where $a \neq 0$.

a) For a parabola in this family with its turning point on the $x$-axis, find $a$ in terms of $b$.

b) If the turning point is at $(4, 0)$, find the values of $a$ and $b$.

Solution:

a) The discriminant is $\Delta = b^2 - 8a$

For the quadratic $ax^2 + bx + c$, the discriminant is $\Delta = b^2 - 4ac$. In this case, $c = 2$.

Since the parabola touches the $x$-axis at its turning point, $\Delta = 0$:

Therefore: $a = \frac{b^2}{8}$

b) The axis of symmetry has equation $x = -\frac{b}{2a}$

Since the turning point is at $(4, 0)$, we know $x = 4$ is the axis of symmetry:

This gives us: $b = -8a$

From part a, we have $a = \frac{b^2}{8}$

Substituting $b = -8a$:

Since $a \neq 0$, we get: $a = \frac{1}{8}$

Substituting into $b = -8a$:

Answer: $a = \frac{1}{8}$ and $b = -1$

Explanation: When a parabola touches the $x$-axis at exactly one point (its turning point), the discriminant equals zero. The axis of symmetry formula $x = -\frac{b}{2a}$ helps us relate the parameters to the turning point's $x$-coordinate.

Worked Example: Using intercept form

A parabola has $x$-axis intercepts $-3$ and $4$, and it passes through the point $(1, 24)$. Find the rule for this parabola.

Solution:

Since two $x$-axis intercepts are given, use the form $y = a(x - e)(x - f)$:

Substitute the point $(1, 24)$:

Answer: The rule is $y = -2(x + 3)(x - 4)$

Explanation: The intercept form is ideal here because we know exactly where the parabola crosses the $x$-axis. We substitute the third point to find the value of $a$.

Worked Example: Using turning point form

The coordinates of the turning point of a parabola are $(2, 6)$ and the parabola passes through the point $(3, 3)$. Find the rule for this parabola.

Solution:

Since the turning point and one other point are given, use the form $y = a(x - h)^2 + k$:

Substitute the point $(3, 3)$:

Answer: The rule is $y = -3(x - 2)^2 + 6$

Explanation: When we know the turning point (vertex) coordinates, the turning point form is most efficient. The values $h = 2$ and $k = 6$ go directly into the equation, and we just need to find $a$ using the additional point.

Worked Example: Using general form with three points

A parabola passes through the points $(1, 4)$, $(0, 5)$ and $(-1, 10)$. Find the rule for this parabola.

Solution:

Since three points are given, use the form $y = ax^2 + bx + c$:

Substitute $(1, 4)$:

Substitute $(0, 5)$:

Substitute $(-1, 10)$:

From equation (2), we know $c = 5$

Substitute into equations (1) and (3):

Add equations (1') and (3'):

Using equation (1'): $-1 = 2 + b$, so $b = -3$

Answer: The rule is $y = 2x^2 - 3x + 5$

Explanation: With three points and no other special information, we use the general form. This creates a system of three equations with three unknowns, which we solve using simultaneous equation techniques.

Worked Example: Four different parabolas

Determine the quadratic rule for each of the following parabolas:

Solution:

a) This parabola passes through $(2, 5)$ with its vertex at the origin.

Use the form $y = ax^2$:

For point $(2, 5)$: $5 = 4a$

Therefore: $a = \frac{5}{4}$

Rule: $y = \frac{5}{4}x^2$

b) This parabola has a vertex at $(0, 3)$ and passes through $(-3, 1)$.

Use the form $y = ax^2 + c$:

From the $y$-intercept: $c = 3$

For point $(-3, 1)$: $1 = a(-3)^2 + 3$

Therefore: $a = -\frac{2}{9}$

Rule: $y = -\frac{2}{9}x^2 + 3$

c) This parabola has $x$-intercepts at $0$ and $3$, and passes through $(-1, 8)$.

Use the form $y = ax(x - 3)$:

For point $(-1, 8)$: $8 = -a(-1 - 3)$

Therefore: $a = 2$

Rule: $y = 2x(x - 3) = 2x^2 - 6x$

d) This parabola has a turning point at $(1, 6)$ and passes through $(0, 8)$.

Use the form $y = a(x - 1)^2 + 6$:

For point $(0, 8)$: $8 = a(0 - 1)^2 + 6$

Therefore: $a = 2$

Rule: $y = 2(x - 1)^2 + 6 = 2x^2 - 4x + 8$

Key insight: The choice of form depends on the features visible in each graph. Look for intercepts, vertices, or other special points to guide your selection.