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

Worked Example: Analyzing a Parametric Cubic Family

Let's examine the family of functions with rules of the form:

where $a$ and $b$ are positive constants with $b > a$.

Finding the derivative

When we differentiate this function with respect to $x$, we get:

This derivative still contains the parameters $a$ and $b$, which means the location of stationary points will depend on these values.

Finding stationary points

Stationary points occur where $f'(x) = 0$. Setting the derivative equal to zero:

This equation has two solutions:

• $x = a$
• $3x - a - 2b = 0$, which gives $x = \frac{a + 2b}{3}$

To find the $y$-coordinates, we substitute these $x$-values back into the original function.

When $x = a$:

When $x = \frac{a + 2b}{3}$:

Therefore, the coordinates of the stationary points are:

Determining the nature of the stationary point

To show that the point $(a, 0)$ is always a local maximum, we examine the sign of $f'(x)$ on either side of $x = a$.

When $x < a$:

Both factors $(x - a)$ and $(3x - a - 2b)$ are negative, so their product is positive: $f'(x) > 0$

The function is increasing as we approach $x = a$ from the left.

When $a < x < \frac{a + 2b}{3}$:

The factor $(x - a)$ is positive, but $(3x - a - 2b)$ is negative, so: $f'(x) < 0$

The function is decreasing as we move away from $x = a$ to the right.

Since the derivative changes from positive to negative at $x = a$, the stationary point at $(a, 0)$ is always a local maximum.

Finding specific parameter values

If we know the stationary points occur at $x = 3$ and $x = 4$, we can find the values of $a$ and $b$.

Since $a < b$, the smaller $x$-value must be $a$:

The larger $x$-value corresponds to the second stationary point:

Substituting $a = 3$:

Worked Example: Effect of Translation on Turning Points

Consider the graph of $y = x^3 - 3x^2$. We want to understand what happens when this graph is translated by $a$ units in the positive $x$-direction and $b$ units in the positive $y$-direction (where $a$ and $b$ are positive constants).

Finding the original turning points

First, we find the turning points of $y = x^3 - 3x^2$.

Taking the derivative:

Setting this equal to zero:

This gives $x = 0$ or $x = 2$.

Substituting back into the original equation:

When $x = 0$: $y = 0$

When $x = 2$: $y = 8 - 12 = -4$

The turning points have coordinates $(0, 0)$ and $(2, -4)$.

Effect of translation on turning points

When a graph is translated by $a$ units horizontally and $b$ units vertically, every point $(x, y)$ on the original graph moves to the point $(x + a, y + b)$ on the translated graph.

Therefore, the turning points of the translated image are:

Notice that the $x$-coordinates increase by $a$ and the $y$-coordinates increase by $b$, exactly as expected for this translation.

Worked Example: Determining Coefficients Using Stationary Point Conditions

A cubic function has the rule $f(x) = ax^3 + bx^2 + cx$ and has a stationary point at $(1, 6)$.

Finding $a$ and $b$ in terms of $c$

Since the point $(1, 6)$ lies on the curve, we know that $f(1) = 6$:

Since $(1, 6)$ is a stationary point, the derivative equals zero at $x = 1$.

The derivative is:

Therefore:

We now have two equations with three unknowns. To express $a$ and $b$ in terms of $c$, we solve these simultaneous equations.

Subtracting equation (2) from equation (1):

Substituting this into equation (1):

Substituting back into equation (3):

Therefore: $a = c - 12$ and $b = 18 - 2c$

Finding the value of $c$ for a specific condition

If the graph has another stationary point at $x = 2$, we can find the value of $c$.

The function rule becomes:

The derivative is:

Since there's a stationary point at $x = 2$, we have $f'(2) = 0$: