Year 10A Function Notation Worksheets & Practice

Worked examples

Year 10A function notation: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Evaluating a function

   A function is defined as $f(x) = 3x^2 - 2x + 1$. Find $f(4)$.

   Show worked solution

   Key idea: To evaluate $f(4)$, substitute $x = 4$ into the rule and simplify.

   1. Write the function: $f(x) = 3x^2 - 2x + 1$
   2. Substitute $x = 4$:
      $f(4) = 3(4)^2 - 2(4) + 1$
   3. Evaluate each term:
      $\begin{aligned} f(4) &= 3 \times 16 - 8 + 1 \\ &= 48 - 8 + 1 \end{aligned}$
   4. Simplify:
      $f(4) = 41$

   Answer: $f(4) = 41$

2. Example 2 · Finding x given f(x)

   A function is defined as $f(x) = 2x^2 - 3x - 9$. Find all values of $x$ for which $f(x) = 0$.

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   Key idea: Set the function equal to zero and solve the resulting equation.

   1. Set up the equation:
      $2x^2 - 3x - 9 = 0$
   2. Find two numbers that multiply to $2 \times (-9) = -18$ and add to $-3$: these are $3$ and $-6$. Split the middle term:
      $2x^2 + 3x - 6x - 9 = 0$
   3. Factor by grouping:
      $\begin{aligned} x(2x + 3) - 3(2x + 3) &= 0 \\ (x - 3)(2x + 3) &= 0 \end{aligned}$
   4. Apply the null factor law:
      $x = 3 \text{ or } x = -\tfrac{3}{2}$

   Answer: $x = 3$ or $x = -\dfrac{3}{2}$

3. Example 3 · Function notation in a real-world model

   A company models its weekly profit (in dollars) from selling $x$ handmade candles using the function $P(x) = -2x^2 + 80x - 300$, where $x$ is the number of candles sold and $x \geq 0$.

   (a) Find $P(10)$ and interpret what it means.

   (b) Find the value of $x$ for which $P(x) = 0$, and interpret the smaller of the two values.

   Show worked solution

   Key idea: Use function notation to substitute values and solve equations; interpret each answer in context.

   Part (a)

   1. Substitute $x = 10$:
      $P(10) = -2(10)^2 + 80(10) - 300$
   2. Evaluate:
      $\begin{aligned} P(10) &= -2 \times 100 + 800 - 300 \\ &= -200 + 800 - 300 \\ &= 300 \end{aligned}$ When the company sells 10 candles, the weekly profit is $300.

   Part (b) 3. Set $P(x) = 0$:
   $-2x^2 + 80x - 300 = 0$ Divide every term by $-2$:
   $x^2 - 40x + 150 = 0$ 4. Use the quadratic formula with $a = 1$, $b = -40$, $c = 150$:
   $x = \frac{40 \pm \sqrt{1600 - 600}}{2} = \frac{40 \pm \sqrt{1000}}{2} = \frac{40 \pm 10\sqrt{10}}{2} = 20 \pm 5\sqrt{10}$ 5. Evaluate both values ($\sqrt{10} \approx 3.162$):
   $x \approx 20 - 15.81 \approx 4.19 \text{ or } x \approx 20 + 15.81 \approx 35.81$ The smaller value $x \approx 4.19$ means the company must sell at least 5 candles before it begins to make a profit (i.e. the break-even point on the way up).

   Answer: (a) $P(10) = 300$; selling 10 candles gives a weekly profit of $300. (b) $x = 20 - 5\sqrt{10} \approx 4.19$ and $x = 20 + 5\sqrt{10} \approx 35.81$; the smaller value is the lower break-even point.

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