Linear and quadratic functions (QCE Math Methods Unit 1)

What this dot point is asking

QCAA wants you to sketch and analyse linear and quadratic functions, find gradient, intercepts and vertex, and solve linear and quadratic equations and inequalities.

Linear functions

Form

$y = mx + c$. Gradient $m$, $y$-intercept $c$.

Gradient from two points

$m = (y_2 - y_1)/(x_2 - x_1)$.

Point-slope form

$y - y_1 = m(x - x_1)$.

$x$-intercept

$x = -c/m$.

Parallel and perpendicular

$m_1 = m_2$ for parallel; $m_1 m_2 = -1$ for perpendicular.

Inequalities

Solve as equations, but reverse the inequality sign when multiplying or dividing by a negative number.

A linear function has a constant gradient, so its graph is a straight line with no turning points. The gradient measures the rate of change of $y$ with respect to $x$, which is the first link to the calculus idea of a derivative introduced in later units.

Quadratic functions

Standard form

$y = ax^2 + bx + c$.

Factored form

$y = a(x - p)(x - q)$. Roots at $p$ and $q$.

Vertex form

$y = a(x - h)^2 + k$. Vertex at $(h, k)$.

Vertex from standard form

$x_v = -b/(2a)$. $y_v = c - b^2/(4a)$.

The discriminant

$\Delta = b^2 - 4ac$.

Quadratic formula

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

Sketching a parabola

1. Identify $a$ (opens up if $a > 0$, down if $a < 0$).
2. Find $y$-intercept (substitute $x = 0$).
3. Find vertex via $-b/(2a)$.
4. Find $x$-intercepts (factor, complete the square, or use the formula).
5. Plot and draw a smooth parabola.

Completing the square

Converting standard form to vertex form reveals the turning point directly. For $y = x^2 + 6x + 1$, halve the coefficient of $x$ (giving $3$), square it ($9$), and add and subtract it: $y = (x^2 + 6x + 9) - 9 + 1 = (x + 3)^2 - 8$. The turning point is $(-3, -8)$. Completing the square is also the derivation behind the quadratic formula and is the method when a question asks for the vertex in exact form.

Solving quadratic inequalities

To solve $ax^2 + bx + c > 0$, first find the roots, then use the parabola's shape. If $a > 0$ the parabola opens upward, so the expression is positive outside the roots and negative between them; if $a < 0$ the reverse holds. For example $x^2 - x - 6 > 0$ factors as $(x - 3)(x + 2) > 0$, with roots $3$ and $-2$; since the parabola opens up, the solution is $x < -2$ or $x > 3$. A quick sketch of the parabola removes any doubt about which region to choose.

In one sentence

Linear functions $y = mx + c$ have gradient $m$ (parallel lines share $m$, perpendicular have $m_1 m_2 = -1$), and quadratics $y = ax^2 + bx + c$ have vertex at $x = -b/(2a)$ and discriminant $\Delta = b^2 - 4ac$ that classifies the roots ($> 0$ two real, $= 0$ one repeated, $< 0$ none).