Linear and quadratic graphs

Linear graphs

• A linear equation produces a straight line:

$$y = mx + c$$

• $m$ — the gradient: positive rises, negative falls, zero is horizontal

• $c$ — the $y$-intercept, where the line crosses the vertical axis

• To sketch a line quickly, plot the $y$-intercept and use the gradient as a step: from $(0, c)$, go 1 right and $m$ up.

• The $x$-intercept is found by setting $y = 0$ and solving.

• A vertical line $x = a$ is not a function, because one $x$-value has many $y$-values. It fails the vertical line test.

Quadratic graphs

• A quadratic produces a parabola — a symmetric U or ∩ shape:

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

• The sign of $a$ decides which way it opens:
  • $a > 0$ — opens upwards, with a minimum
  • $a < 0$ — opens downwards, with a maximum
• The size of $a$ decides how narrow it is. A large $|a|$ gives a steep, narrow parabola.
• $c$ is the $y$-intercept, since setting $x = 0$ leaves only $c$.

The three forms of a quadratic

• Each form displays a different feature immediately. Choosing the right one is most of the skill.

• General form: $y = ax^2 + bx + c$

  • Shows the $y$-intercept ($c$) at a glance.

• Factorised form: $y = a(x - p)(x - q)$

  • Shows the $x$-intercepts at $x = p$ and $x = q$.
  • The axis of symmetry is midway between them: $x = \dfrac{p+q}{2}$.

• Completed-square (vertex) form: $y = a(x - h)^2 + k$

  • Shows the vertex directly at $(h, k)$.
  • Watch the sign: $y = 2(x - 3)^2 + 1$ has its vertex at $(3, 1)$, and $y = 2(x + 3)^2 + 1$ at $(-3, 1)$.

Finding the vertex

• From factorised form: average the two roots to get the $x$-coordinate, then substitute for $y$.
• From general form: the axis of symmetry is

$$x = -\frac{b}{2a}$$

then substitute to find $y$.

• From completed-square form: read it straight off.

• Worked through for $y = x^2 - 6x + 5$:

  • Factorised: $y = (x-1)(x-5)$, so the roots are 1 and 5
  • Axis of symmetry: $x = \frac{1+5}{2} = 3$
  • Vertex: $y = 9 - 18 + 5 = -4$, so $(3, -4)$
  • Or by completing the square: $y = (x-3)^2 - 4$, giving the vertex directly ✓

How many $x$-intercepts

• The discriminant $\Delta = b^2 - 4ac$ counts the crossings:

  • $\Delta > 0$ — two $x$-intercepts
  • $\Delta = 0$ — one; the parabola touches the axis at its vertex
  • $\Delta < 0$ — none; the parabola misses the axis entirely

• Note that a parabola with no $x$-intercepts still has a $y$-intercept and a vertex — it simply sits entirely above or entirely below the axis.

Domain and range

• Domain — the set of $x$-values the function accepts.

• Range — the set of $y$-values it produces.

• For a line (non-vertical): domain is all real $x$; range is all real $y$.

• For a parabola: domain is all real $x$; the range is limited by the vertex:

  • $a > 0$: range is $y \geq k$
  • $a < 0$: range is $y \leq k$

• The vertex is what makes the range interesting. A parabola cannot produce values beyond its turning point, which is why the range is a half-line rather than everything.