Quadratic functions and parabolas (VCE Maths Methods Unit 1)

What this dot point is asking

VCAA wants you to sketch and analyse quadratic functions in their three standard forms, find vertex and intercepts, and use the discriminant to classify roots.

The three forms

Standard form

$y = ax^2 + bx + c$. Coefficient $a$ controls opening (up if $a > 0$, down if $a < 0$) and steepness. $c$ is the $y$-intercept.

Factored form

$y = a(x - p)(x - q)$. Roots are at $x = p$ and $x = q$. Useful when roots are known.

Turning-point (vertex) form

$y = a(x - h)^2 + k$. Vertex at $(h, k)$. Axis of symmetry $x = h$. Useful when the vertex is known.

Convert between forms by expanding (factored or vertex to standard) or completing the square (standard to vertex).

Vertex from standard form

The discriminant

Classifies the roots of $ax^2 + bx + c = 0$:

Quadratic formula

Completing the square

To move from standard form $y = ax^2 + bx + c$ to turning-point form $y = a(x - h)^2 + k$, complete the square. Factor $a$ from the first two terms, add and subtract the square of half the new linear coefficient, then tidy. For $y = 2x^2 - 8x + 5$: $y = 2(x^2 - 4x) + 5 = 2((x - 2)^2 - 4) + 5 = 2(x - 2)^2 - 3$, confirming the vertex $(2, -3)$. Completing the square is the technique behind both the vertex formula and the quadratic formula, and is examinable as a method in its own right.

Finding a quadratic from conditions

A parabola is determined by three independent pieces of information, so VCAA problems often give the vertex plus a point, two intercepts plus a point, or three general points, and ask you to construct the rule. With the vertex $(h, k)$ known, write $y = a(x - h)^2 + k$ and use a second point to solve for $a$. With the roots $p$ and $q$ known, write $y = a(x - p)(x - q)$ and use a point for $a$. Choosing the form that matches the given features keeps the algebra short.

In one sentence

Quadratics $y = ax^2 + bx + c$ have vertex at $x_v = -b/(2a)$, classify their roots through the discriminant $\Delta = b^2 - 4ac$ ($> 0$ two real, $= 0$ one repeated, $< 0$ none), and can be written in standard, factored ($y = a(x - p)(x - q)$) or vertex ($y = a(x - h)^2 + k$) form depending on which features are known.

Examples in context

Example 1. Maximum revenue. A stall's daily revenue is modelled by $R = -2p^2 + 24p$ dollars, where $p$ is the price in dollars. This is a downward parabola, so the maximum is at the vertex: $p_v = -\frac{b}{2a} = -\frac{24}{2(-2)} = 6$. The maximum revenue is $R(6) = -2(36) + 24(6) = -72 + 144 = \$72$, achieved at a price of $\$6$.

Example 2. Projectile landing point. A ball's height is $h = -5t^2 + 20t$ metres for time $t$ seconds. It lands when $h = 0$: $-5t^2 + 20t = -5t(t - 4) = 0$, giving $t = 0$ (launch) and $t = 4$ s (landing). The peak is at the vertex $t = -\frac{20}{2(-5)} = 2$ s, where $h = -5(4) + 40 = 20$ m.

Try this

Q1. For $y = x^2 - 6x + 5$, find the vertex and the $x$-intercepts. [3 marks]

• Cue. $x_v = 3$, $y_v = 9 - 18 + 5 = -4$, vertex $(3, -4)$; factor $(x - 1)(x - 5)$, intercepts $x = 1, 5$.

Q2. Find the values of $k$ for which $x^2 + kx + 9 = 0$ has exactly one solution. [3 marks]

• Cue. $\Delta = k^2 - 36 = 0 \Rightarrow k = \pm 6$.

Q3. A fountain's water arc is $y = -0.5(x - 4)^2 + 8$ (metres). (a) State the maximum height and where it occurs. (b) Find where the water lands ($y = 0$). [2+2 marks]

• Cue. (a) Max $8$ m at $x = 4$. (b) $(x - 4)^2 = 16 \Rightarrow x = 0$ or $x = 8$; lands at $x = 8$ m.