Quadratic Equations & Graphs

Quadratics describe situations where the variable is squared, so their graphs curve instead of forming straight lines. In CSEC Mathematics, quadratics connect algebra and graphs: you may be asked to solve an equation, identify roots, find a turning point, or interpret the shape of a parabola.

The important idea is that each method reveals something different. Factorising shows where the graph crosses the $x$-axis, completing the square shows the turning point, and the quadratic formula works even when simple factorising does not. When a question carries reasoning marks, explain why the method you chose fits the form of the quadratic.

Solving Quadratics by Factorization¶

Factorisation is the quickest method when the quadratic breaks neatly into brackets. Once the product equals zero, at least one bracket must be zero. This is called the zero product property, and it is the reason the bracketed factors become separate linear equations.

If $ax^2 + bx + c = 0$ factors to $(px + q)(rx + s) = 0$, then: $px + q = 0 \text{ or } rx + s = 0$

Solving by Completing the Square¶

Completing the square is slower than factorising, but it gives more information. It rewrites a quadratic so you can see its turning point clearly. This is especially useful for graph questions and for quadratics that do not factor nicely.

Express $ax^2 + bx + c$ as a perfect square plus a constant. This method also helps find the vertex (minimum or maximum point).

General Method:

For $ax^2 + bx + c = 0$:

$a\left(x^2 + \frac{b}{a}x\right) + c = 0$

$a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2\right) + c - a\left(\frac{b}{2a}\right)^2 = 0$

$a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} = 0$

This gives vertex form: $a(x - h)^2 + k = 0$

Where:

• $h = -\frac{b}{2a}$ (x-coordinate of vertex)
• $k = c - \frac{b^2}{4a}$ (y-coordinate of vertex)
• Vertex $(h, k)$ is the minimum (if $a > 0$) or maximum (if $a < 0$)

Example 1: Simple Case ($a = 1$)¶

In this first case, the coefficient of $x^2$ is 1, so you can complete the square directly. The number added to the left must also be added to the right because the equation must stay balanced.

$x^2 + 6x - 7 = 0$

Step 1: Move constant to right side $x^2 + 6x = 7$

Step 2: Take half the x-coefficient $\frac{b}{2} = \frac{6}{2} = 3$

Step 3: Square it $\left(\frac{6}{2}\right)^2 = 9$

Step 4: Add to both sides $x^2 + 6x + 9 = 7 + 9$

Step 5: Factor left as perfect square $(x + 3)^2 = 16$

Step 6: Solve $x + 3 = \pm 4$ $x = -3 + 4 = 1 \text{ or } x = -3 - 4 = -7$

Vertex (from form): $(x + 3)^2 = 16$ means vertex at $x = -3$, and when $x = -3$: $(0)^2 = 16$ gives $y = 16$... wait, that's not right at 0. At $x = -3$: $y = (-3)^2 + 6(-3) - 7 = 9 - 18 - 7 = -16$

So vertex is at $(-3, -16)$ ← minimum point

Check: $1^2 + 6(1) - 7 = 1 + 6 - 7 = 0$ ✓

Example 2: General Case ($a ≠ 1$)¶

When $a$ is not 1, factor it out of the $x^2$ and $x$ terms first. This prevents you from adding the wrong amount to the expression.

$2x^2 + 8x - 10 = 0$

Step 1: Factor out leading coefficient from first two terms $2\left(x^2 + 4x\right) - 10 = 0$

Step 2: Complete the square inside the brackets

• Half the x-coefficient: $\frac{4}{2} = 2$
• Square it: $2^2 = 4$

$2\left(x^2 + 4x + 4\right) - 10 - 2(4) = 0$

Note: We subtract $2 \times 4 = 8$ because we added $2 \times 4$ inside

Step 3: Factor as perfect square $2(x + 2)^2 - 18 = 0$

Step 4: Rearrange to find vertex $2(x + 2)^2 = 18$ $(x + 2)^2 = 9$

Step 5: Solve $x + 2 = \pm 3$ $x = -2 + 3 = 1 \text{ or } x = -2 - 3 = -5$

Vertex form: $2(x + 2)^2 - 18 = 0$ can be written as $2(x - (-2))^2 + (-18) = 0$

So vertex at $(-2, -18)$ ← minimum point (since $a = 2 > 0$)

Check: $2(1)^2 + 8(1) - 10 = 2 + 8 - 10 = 0$ ✓

Finding Min/Max Point Directly¶

Once in vertex form $a(x - h)^2 + k$, the vertex is immediately $(h, k)$:

Quadratic Formula¶

The quadratic formula is the most dependable solving method because it works for every quadratic in the form $ax^2 + bx + c = 0$. It is useful when factorising is difficult or impossible.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For equation $ax^2 + bx + c = 0$

Discriminant¶

The discriminant is the part under the square root in the quadratic formula. Before solving fully, it tells you how many real answers to expect, which helps you check whether your final result makes sense.

The discriminant $\Delta = b^2 - 4ac$ tells us about the solutions:

• $\Delta > 0$: Two distinct real solutions
• $\Delta = 0$: One repeated real solution
• $\Delta < 0$: No real solutions (complex solutions)

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Part 12: Quadratic Graphs¶

Parabolas¶

A parabola is the visual form of a quadratic. The roots, intercepts, vertex, and axis of symmetry are all different ways of describing the same curve.

A quadratic function $y = ax^2 + bx + c$ graphs as a parabola.

Properties:

• If $a > 0$: opens upward (U-shape), has minimum point
• If $a < 0$: opens downward (∩-shape), has maximum point
• Vertex = turning point (minimum or maximum)

Vertex Form¶

Vertex form is designed for graph interpretation. It tells you the turning point immediately, so it is often more useful than standard form when the question asks for a maximum, minimum, or axis of symmetry.

$y = a(x-h)^2 + k$

Where $(h, k)$ is the vertex.

Converting Standard Form to Vertex Form¶

Use completing the square:

$y = ax^2 + bx + c$ $y = a\left(x^2 + \frac{b}{a}x\right) + c$ $y = a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2\right) + c - a\left(\frac{b}{2a}\right)^2$ $y = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$

Vertex: $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$

Quick Vertex Formula¶

The quick formula is a shortcut for the $x$-coordinate of the turning point. It does not replace understanding the graph; after finding $x$, you still substitute into the original equation to find the matching $y$ value.

If you don't want to complete the square every time:

$x_{\text{vertex}} = -\frac{b}{2a}$

Then substitute back to find $y_{\text{vertex}}$:

$y_{\text{vertex}} = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c$