Master Quadratic Equations and Polynomials - Study Notes

Getting Started with Quadratics

This is your complete guide to mastering quadratics and polynomials at Higher level. You'll learn the key techniques that turn up in exams year after year, with plenty of worked examples to build your confidence.

The skills covered here – completing the square, sketching graphs, and working with polynomials – are fundamental building blocks for calculus and other advanced topics. Master these now and you'll find the rest of Higher Maths much more manageable.

💡 Study Tip: These techniques follow logical patterns. Once you spot the patterns, you'll find quadratics surprisingly straightforward!

Completing the Square

Completing the square transforms any quadratic into the neat form y = $x - a$² + b, which makes sketching graphs dead easy. The method depends on whether your x coefficient is even, odd, or if there's a number in front of x².

For even x coefficients like x² + 8x + 9, you halve the x coefficient (getting 4), square it in brackets: $x + 4$² + 9 - 16 = $x + 4$² - 7. With odd coefficients like x² + 5x - 3, you get fractions: $x + 5/2$² - 37/4.

When there's a coefficient in front of x² $like 2x² + 12x + 10$, factor it out first: 2$x² + 6x$ + 10, then complete the square inside the brackets. This gives you 2$x + 3$² - 8.

💡 Remember: Always subtract the square of your half-coefficient – that's where most mistakes happen!

More Completing the Square Practice

The examples on this page show you how to handle fractional coefficients and negative x² terms. Don't panic when you see fractions – the method stays exactly the same.

For expressions like x² + x + 3, you get $x + 1/2$² + 11/4 after subtracting $1/2$². When dealing with negative x² coefficients like in 6 - x - 2x², rearrange to -2x² - x + 6 first, then factor out the -2.

The key is working systematically through each step. Factor out coefficients, complete the square in brackets, then simplify. With practice, you'll spot these patterns instantly.

💡 Pro Tip: Check your answer by expanding back out – it should match your original expression!

Sketching Quadratic Graphs

Sketching parabolas becomes straightforward once you know what information to find. Every quadratic graph needs five key features: where it crosses both axes, the turning point coordinates, the axis of symmetry, and whether it opens upward or downward.

If a > 0, your parabola has a minimum turning point (happy face). If a < 0, it has a maximum turning point (sad face). Find the x-intercepts by setting y = 0, and the y-intercept by setting x = 0.

The axis of symmetry runs exactly halfway between the roots. For y = x + 3$$x + 5, the roots are x = -3 and x = -5, so the axis of symmetry is x = -4. Substitute this back to find the turning point.

💡 Quick Check: Your turning point should always lie on the axis of symmetry!

Using Completed Square Form for Sketching

When your quadratic is in completed square form, sketching becomes incredibly quick. For y = 2$x - 2$² + 1, you can immediately read off the turning point as (2, 1) and the axis of symmetry as x = 2.

The coefficient in front tells you the shape. Since we have +2, this parabola opens upward with a minimum turning point. For y = -$x - 3$² + 16, the negative sign means it opens downward with a maximum at (3, 16).

Always find where the graph crosses the y-axis by substituting x = 0. This gives you enough information to sketch an accurate graph that'll earn you full marks in exams.

💡 Time Saver: Completed square form gives you the turning point instantly – no extra calculations needed!

Finding Quadratic Equations from Graphs

When you're given a graph and need to find its equation, you'll use the form y = kx - a$$x - b where a and b are the x-intercepts. The tricky bit is finding the value of k.

Use any point on the curve (often given or easy to read off) to find k. If your parabola crosses the x-axis at x = 4 and x = -1, and passes through $0, -12$, then -12 = k$-4$(1), so k = 3.

Check your answer makes sense by substituting another point from the graph. This method works for any quadratic, whether it opens upward or downward.

💡 Exam Tip: The y-intercept is usually the easiest extra point to use for finding k!

Solving Quadratic Inequalities

Quadratic inequalities tell you when a quadratic expression is positive or negative. The secret is sketching the graph and reading off where it's above or below the x-axis.

For x² + 4x - 12 > 0, first solve x² + 4x - 12 = 0 to get x = -6 and x = 2. Since the coefficient of x² is positive, this parabola opens upward. The expression is positive (above the x-axis) when x < -6 or x > 2.

For inequalities with negative x² coefficients, the parabola opens downward. So for 7 + 6x - x² < 0, you want the regions where this downward-opening parabola sits below the x-axis.

💡 Visual Trick: Above the x-axis = positive, below the x-axis = negative. Let the graph do the work!

Tangents to Parabolas

The discriminant $b² - 4ac$ tells you exactly how a straight line interacts with a parabola. When you substitute a line's equation into a parabola's equation, you get a quadratic whose discriminant reveals everything.

If b² - 4ac > 0, there are two intersection points. If b² - 4ac = 0, there's exactly one point of contact – the line is tangent to the curve. If b² - 4ac < 0, there are no intersections at all.

To find tangent equations, set up your equation with the line y = mx + c touching the parabola, then force the discriminant to equal zero. This gives you the value of any unknown coefficients.

💡 Key Insight: A tangent just touches the curve at exactly one point – that's why the discriminant equals zero!

Finding Specific Tangents

When you need to find tangent equations with specific properties, use the discriminant method systematically. For a tangent to y = x² + 1 with gradient 2, set up y = 2x + c and solve for c using b² - 4ac = 0.

For tangents from a specific point like $0, -2$ to curve y = 8x², you'll get two possible gradients. Set up y = mx - 2, substitute into the curve equation, then solve m² = 64 to get m = ±8.

This gives you two tangent lines: y = 8x - 2 and y = -8x - 2. Always check your answers by verifying they pass through the given point.

💡 Double Check: From any external point, you can usually draw exactly two tangents to a parabola!

Polynomial Basics

Polynomials are expressions with different powers of the same variable, like 2x⁴ - 3x³ + x² - 5. The degree is simply the highest power – so this polynomial has degree 4.

A root of a polynomial f$x$ is any value where f$x$ = 0. To check if a number is a root, substitute it in and see if you get zero. For f$x$ = x³ - 4x² + x + 6, trying x = 2 gives f(2) = 8 - 16 + 2 + 6 = 0, so x = 2 is definitely a root.

Finding roots by testing values is often the first step in factoring polynomials. Once you find one root, you can use polynomial division to find the others.

💡 Exam Strategy: Try simple values like ±1, ±2, ±3 first – they're the most likely roots in exam questions!