Year 10 and 10A Polynomials, functions and graphs: free practice test with answers

Sample polynomials, functions and graphs questions with answers

Try each one before you open the answer.

1. Functions and their notation

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

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   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. Introducing polynomials

   For the polynomial $P(x) = 5x^{4} - 3x^{2} + 2x - 9$:

   (a) State the degree.

   (b) State the coefficient of $x^{3}$.

   (c) Find $P(1)$.

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   Key idea: The degree is the highest power; a missing power has coefficient 0.

   1. The highest power is $x^{4}$, so the degree is $4$
   2. There is no $x^{3}$ term, so its coefficient is $0$
   3. $P(1) = 5 - 3 + 2 - 9 = -5$

   Answer: (a) $4$ (b) $0$ (c) $-5$

3. Expanding and simplifying polynomials

   Expand and simplify $(x^2 - 3x + 2)(2x + 5) - (x - 1)^3$.

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   Key idea: Expand each product fully, then subtract, taking care to change every sign of the second expansion.

   1. $(x^2 - 3x + 2)(2x + 5) = 2x^3 + 5x^2 - 6x^2 - 15x + 4x + 10 = 2x^3 - x^2 - 11x + 10$
   2. $(x - 1)^3 = x^3 - 3x^2 + 3x - 1$
   3. Subtract: $2x^3 - x^2 - 11x + 10 - x^3 + 3x^2 - 3x + 1$
   4. Collect like terms: $x^3 + 2x^2 - 14x + 11$

   Answer: $x^3 + 2x^2 - 14x + 11$

4. The remainder and factor theorems

   Find the remainder when $P(x) = 2x^3 - 5x^2 + x + 3$ is divided by $(x - 2)$.

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   Key idea: The remainder theorem states that when $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.

   1. The divisor is $(x - 2)$, so $a = 2$.
   2. Substitute $x = 2$ into $P(x)$:\

   $$P(2) = 2(2)^3 - 5(2)^2 + (2) + 3$$

   3. Evaluate each term:\

   $$\begin{aligned} P(2) &= 2(8) - 5(4) + 2 + 3 \\ &= 16 - 20 + 2 + 3 \end{aligned}$$

   4. Simplify:\

   $$P(2) = 1$$

   Answer: The remainder is $1$.

5. Solving polynomial equations

   A storage company designs an open-top rectangular box by cutting squares of side length $x$ cm from each corner of a 12 cm by 18 cm rectangular sheet of cardboard, then folding up the sides. The volume of the box is given by $V(x) = x(12 - 2x)(18 - 2x)$. The company needs a volume of 160 cm³. Show that this requirement leads to the equation $x^3 - 15x^2 + 54x - 40 = 0$, then find all valid values of $x$.

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   Key idea: Expand the volume expression and rearrange to form a cubic equation equal to zero, then use the factor theorem and long division to fully factorise and solve.

   1. Expand $V(x) = x(12 - 2x)(18 - 2x)$ and set it equal to 160:\

   $$\begin{aligned} (12 - 2x)(18 - 2x) &= 216 - 24x - 36x + 4x^2 = 4x^2 - 60x + 216 \\ V(x) &= x(4x^2 - 60x + 216) = 4x^3 - 60x^2 + 216x \end{aligned}$$

   Setting $V(x) = 160$: $4x^3 - 60x^2 + 216x = 160$ Divide everything by 4: $x^3 - 15x^2 + 54x - 40 = 0$ ✓ 2. Test integer factors of 40. Try $x = 1$:\

   $$1 - 15 + 54 - 40 = 0 \checkmark$$

   So $(x - 1)$ is a factor. 3. Divide $x^3 - 15x^2 + 54x - 40$ by $(x - 1)$:\

   $$x^3 - 15x^2 + 54x - 40 = (x - 1)(x^2 - 14x + 40)$$

   4. Factorise $x^2 - 14x + 40 = (x - 4)(x - 10)$, so:\

   $$(x-1)(x-4)(x-10) = 0 \implies x = 1, \; x = 4, \; x = 10$$

   5. Check validity: the shorter side of the sheet is 12 cm, so the cut must satisfy $2x < 12$, meaning $x < 6$. Also $x > 0$. So $x = 10$ is rejected.

   Answer: The valid values are $x = 1$ cm and $x = 4$ cm.

6. Graphing cubic functions of the form y = a(x − h)^3 + k

   For the cubic $y = (x - 2)^3 + 1$, state the point of inflection and find the $y$-intercept and the $x$-intercept.

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   Key idea: For $y = a(x - h)^3 + k$, the point of inflection is $(h, k)$.

   1. Point of inflection: $(2, 1)$.
   2. $y$-intercept: $y = (0 - 2)^3 + 1 = -8 + 1 = -7$, giving $(0, -7)$.
   3. $x$-intercept: $(x - 2)^3 = -1$, so $x - 2 = \sqrt[3]{-1} = -1$ and $x = 1$, giving $(1, 0)$.

   Answer: Point of inflection $(2, 1)$; $y$-intercept $(0, -7)$; $x$-intercept $(1, 0)$

7. Graphs of circles

   The circle $x^2 + y^2 = 49$ has its centre at the origin. What is its radius?

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   Key idea: A circle with centre $(0, 0)$ and radius $r$ has equation $x^2 + y^2 = r^2$.

   1. Compare $x^2 + y^2 = 49$ with the standard form $x^2 + y^2 = r^2$.
   2. Identify $r^2 = 49$.
   3. Take the positive square root: $r = \sqrt{49} = 7$.
   4. Check: a point on the circle such as $(7, 0)$ satisfies $7^2 + 0^2 = 49$. ✓

   Answer: $7$ units

8. Hyperbolic functions and their graphs

   Consider the hyperbola $y = \frac{12}{x}$.

   (a) Find $y$ when $x = 4$.

   (b) Find $x$ when $y = -6$.

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   Key idea: Substitute, then solve.

   1. $y = \frac{12}{4} = 3$
   2. $-6 = \frac{12}{x}$, so $x = \frac{12}{-6} = -2$

   Answer: (a) $3$ (b) $-2$

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 10A function notation worksheet
• Year 10A polynomials and factor theorem worksheet
• Year 10 graphs of parabolas, circles, hyperbolas and exponentials worksheet