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VICMath MethodsSyllabus dot point

How are quadratic functions analysed?

Sketch and analyse quadratic functions in standard, factored and turning-point form, including finding vertex, axis of symmetry, intercepts and using the discriminant to classify roots

A focused answer to the VCE Maths Methods Unit 1 dot point on quadratic functions. Sketches $y = ax^2 + bx + c$, converts between forms, finds the vertex from $x = -b/(2a)$, applies the discriminant $b^2 - 4ac$, and works the VCAA SAC-style turning-point and roots problem.

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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=ax2+bx+cy = ax^2 + bx + c. Coefficient aa controls opening (up if a>0a > 0, down if a<0a < 0) and steepness. cc is the yy-intercept.

Factored form. y=a(xβˆ’p)(xβˆ’q)y = a(x - p)(x - q). Roots are at x=px = p and x=qx = q. Useful when roots are known.

Turning-point (vertex) form. y=a(xβˆ’h)2+ky = a(x - h)^2 + k. Vertex at (h,k)(h, k). Axis of symmetry x=hx = 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

xv=βˆ’b2a,yv=cβˆ’b24ax_v = -\frac{b}{2a}, \quad y_v = c - \frac{b^2}{4a}

The discriminant

Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac

Classifies the roots of ax2+bx+c=0ax^2 + bx + c = 0:

IMATH_16 Roots Parabola
IMATH_17 Two real distinct Crosses xx-axis twice
IMATH_19 One real repeated Touches xx-axis (vertex on it)
IMATH_21 No real roots Does not touch xx-axis

Quadratic formula

x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Worked example

Sketch y=βˆ’(xβˆ’2)2+9y = -(x - 2)^2 + 9.

Vertex form. Vertex at (2,9)(2, 9). a=βˆ’1<0a = -1 < 0, so opens downward.

yy-intercept: y(0)=βˆ’(4)+9=5y(0) = -(4) + 9 = 5.

xx-intercepts: βˆ’(xβˆ’2)2+9=0β€…β€ŠβŸΉβ€…β€Š(xβˆ’2)2=9β€…β€ŠβŸΉβ€…β€Šxβˆ’2=Β±3β€…β€ŠβŸΉβ€…β€Šx=5-(x-2)^2 + 9 = 0 \implies (x-2)^2 = 9 \implies x - 2 = \pm 3 \implies x = 5 or x=βˆ’1x = -1.

Common traps

Sign of bb in xv=βˆ’b/(2a)x_v = -b/(2a). A common slip is to forget the minus sign.

Forgetting that vertex form's hh has opposite sign to the bracket. (xβˆ’3)2(x - 3)^2 has h=+3h = +3, not βˆ’3-3.

Mixing up aa with the leading coefficient. In y=a(xβˆ’h)2+ky = a(x - h)^2 + k, aa is the same coefficient as in standard form ax2+bx+ca x^2 + bx + c.

Treating no-real-roots as no solution. The quadratic has complex roots; they just do not appear on the real-number axis.

In one sentence

Quadratics y=ax2+bx+cy = ax^2 + bx + c have vertex at xv=βˆ’b/(2a)x_v = -b/(2a), classify their roots through the discriminant Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac (>0> 0 two real, =0= 0 one repeated, <0< 0 none), and can be written in standard, factored (y=a(xβˆ’p)(xβˆ’q)y = a(x - p)(x - q)) or vertex (y=a(xβˆ’h)2+ky = a(x - h)^2 + k) form depending on which features are known.

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksFor $f(x) = 2x^2 - 8x + 5$, find (a) the vertex, (b) the discriminant, (c) the exact $x$-intercepts.
Show worked answer β†’

(a) Vertex. xv=βˆ’b/(2a)=8/4=2x_v = -b/(2a) = 8/4 = 2. yv=2(4)βˆ’8(2)+5=8βˆ’16+5=βˆ’3y_v = 2(4) - 8(2) + 5 = 8 - 16 + 5 = -3.

Vertex: (2,βˆ’3)(2, -3).

(b) Discriminant. Ξ”=b2βˆ’4ac=64βˆ’4(2)(5)=64βˆ’40=24>0\Delta = b^2 - 4ac = 64 - 4(2)(5) = 64 - 40 = 24 > 0.

Two real distinct roots.

(c) xx-intercepts. x=8Β±244=8Β±264=4Β±62=2Β±62x = \dfrac{8 \pm \sqrt{24}}{4} = \dfrac{8 \pm 2\sqrt{6}}{4} = \dfrac{4 \pm \sqrt{6}}{2} = 2 \pm \frac{\sqrt 6}{2}.

Markers reward the vertex formula, discriminant evaluation, and simplification of the quadratic-formula surd.

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