Practice Problems for Converting Quadratics to Vertex Form

Students preparing for the HSC in New South Wales, the VCE in Victoria, or the QCE in Queensland spend countless hours refining techniques of algebra. Converting a quadratic equation from standard form into vertex form is one of those techniques that separates confident students from those who struggle under exam pressure. The vertex form y = a(x - h)² + k reveals the turning point of a parabola at a glance, which makes sketching graphs, solving optimisation problems, and interpreting quadratic functions much easier.

The worked problems below start with gentle warm-ups and gradually introduce coefficients other than one, fractions, and negative values. Each example is solved step-by-step so the method stays visible, and the final set of problems links the algebra to situations an Australian student might actually meet in a worded question.

Understanding Vertex Form

Vertex form is written as y = a(x - h)² + k. The value of a controls how stretched or compressed the parabola is and whether it opens upward or downward. The numbers h and k are the coordinates of the vertex, written as (h, k). Finding h and k is the goal of every conversion, so it helps to remember that the sign inside the bracket is the opposite of what most students expect.

If a is positive, the parabola opens upward and the vertex is a minimum. If a is negative, it opens downward and the vertex is a maximum. This single piece of information answers many of the optimisation questions that appear in Year 11 and Year 12 maths courses, from maximising the area of a fenced enclosure to finding the peak height of a thrown ball.

The Completing the Square Method

Completing the square is the reliable method for moving from y = ax² + bx + c into vertex form. The procedure works for any quadratic, including those with fractions or a leading coefficient other than one. The core idea is to rewrite the squared term and the linear term as a perfect square, then adjust the constant so the equation stays balanced.

For a quadratic y = ax² + bx + c, the steps are: factor a from the first two terms, halve the new coefficient of x and square it, add and subtract that squared value inside the bracket, then simplify the constants. After factoring the perfect square, the value that stays outside the bracket is k. The half-and-square trick becomes second nature with repetition, which is exactly what the practice problems below are designed to build.

Practice Set: Simple Quadratics

Start with a quadratic that has a leading coefficient of one. Convert y = x² + 6x + 8 into vertex form.

Factor the constant from the linear coefficient: take half of 6, which is 3, then square it to get 9. Write y = (x² + 6x + 9) - 9 + 8. The bracket is now a perfect square: (x + 3)². Combine the constants: -9 + 8 = -1. The vertex form is y = (x + 3)² - 1, so the vertex sits at (-3, -1).

Try a second one with a negative linear term. Convert y = x² - 10x + 21. Half of -10 is -5, and (-5)² = 25. The equation becomes y = (x² - 10x + 25) - 25 + 21, which simplifies to y = (x - 5)² - 4. The vertex is at (5, -4). Sketching this quickly shows the parabola crossing the x-axis at 3 and 7, since the roots are symmetric about the axis of symmetry x = 5.

Practice Set: Coefficients Greater Than One

When a is not one, the steps gain an extra line. Convert y = 2x² - 8x + 3.

Factor 2 from the first two terms: y = 2(x² - 4x) + 3. Halve -4 to get -2, square it to get 4. Add and subtract 4 inside the bracket: y = 2(x² - 4x + 4 - 4) + 3. Distribute the 2: y = 2(x - 2)² - 8 + 3. The final vertex form is y = 2(x - 2)² - 5, giving a vertex at (2, -5). Because a = 2 is positive, the minimum value of y is -5.

Now try y = -x² + 4x - 1. Factor -1 from the first two terms to keep the bracket tidy: y = -(x² - 4x) - 1. Half of -4 is -2, squared gives 4. Rewrite as y = -(x² - 4x + 4 - 4) - 1, then y = -(x - 2)² + 4 - 1, which becomes y = -(x - 2)² + 3. The vertex is at (2, 3) and since a is negative, this is a maximum value of 3.

Practice Set: Fractions and Decimals

Quadratics with fractional coefficients appear in many NAPLAN-style questions and are common in preliminary courses. Convert y = x² + 3x + 1.

Half of 3 is 1.5, and 1.5² = 2.25. Write y = (x² + 3x + 2.25) - 2.25 + 1, which gives y = (x + 1.5)² - 1.25. The vertex is at (-1.5, -1.25). Fractions can also be handled with the same method: for y = x² + (5/2)x + 1, half of 5/2 is 5/4, and (5/4)² = 25/16. The result is y = (x + 5/4)² - 25/16 + 1 = (x + 5/4)² - 9/16.

A handy sanity check: the axis of symmetry x = -b/(2a) should match the h value inside the bracket. For the first example, h = -3/2 = -1.5, confirming the work.

Real-World Practice Problems

Worded questions often combine vertex form with practical settings. A common problem on VCE papers asks for the maximum area of a rectangular pen built against a barn, using a fixed length of fencing.

Suppose a farmer near Bendigo has 40 metres of fencing for three sides of a rectangular enclosure. If the width is x metres, the length is 40 - 2x, and the area is A = x(40 - 2x) = -2x² + 40x. Converting to vertex form: factor -2, giving A = -2(x² - 20x). Half of -20 is -10, squared is 100. A = -2(x² - 20x + 100 - 100) = -2(x - 10)² + 200. The vertex is at (10, 200), so the maximum area is 200 m² when the width is 10 m and the length is 20 m.

Another problem links to motion. A ball thrown upwards from the top of a 50 m cliff on the Sydney coastline has height h = -5t² + 20t + 50 after t seconds. Converting: h = -5(t² - 4t) + 50 = -5(t² - 4t + 4 - 4) + 50 = -5(t - 2)² + 20 + 50 = -5(t - 2)² + 70. The ball reaches a maximum height of 70 m after 2 seconds.

Recommended Practice Habits

• Rewrite every vertex form back into standard form to check the answer matches the original equation.
• Always confirm the axis of symmetry using x = -b/(2a) when the coefficient a is not one.
• Practise a handful of fractional examples each week so the half-and-square step feels automatic in exams.
• Sketch the parabola after each conversion and mark the vertex, y-intercept, and roots to build graphical intuition.
• Work through past ATAR or HSC papers that include optimisation questions, since worded conversions are tested regularly.

The most reliable way to master converting quadratics into vertex form is repeated, varied practice. Work the warm-ups until the half-and-square step feels automatic, then move on to coefficients other than one, and finish with worded problems that require interpretation rather than just arithmetic. After a focused week of daily practice, the method becomes a single fluid motion that frees attention for the harder reasoning parts of any exam question.