← Unit 1: Algebra, statistics and functions

QLDMath MethodsSyllabus dot point

How are linear and quadratic functions analysed?

Sketch and analyse linear and quadratic functions, finding gradient, intercepts, vertex and discriminant, and solving linear and quadratic equations and inequalities

A focused answer to the QCE Math Methods Unit 1 dot point on linear and quadratic functions. Finds gradient, intercepts and parallel/perpendicular relationships for linear functions; converts between standard, factored and vertex form and uses the discriminant for quadratics.

Generated by Claude OpusReviewed by Better Tuition Academy5 min answer

Have a quick question? Jump to the Q&A page

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+cy = mx + c. Gradient mm, yy-intercept cc.

Gradient from two points. m=(y2βˆ’y1)/(x2βˆ’x1)m = (y_2 - y_1)/(x_2 - x_1).

Point-slope form. yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1).

**xx-intercept.** x=βˆ’c/mx = -c/m.

Parallel and perpendicular. m1=m2m_1 = m_2 for parallel; m1m2=βˆ’1m_1 m_2 = -1 for perpendicular.

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

Quadratic functions

Standard form. y=ax2+bx+cy = ax^2 + bx + c.

Factored form. y=a(xβˆ’p)(xβˆ’q)y = a(x - p)(x - q). Roots at pp and qq.

Vertex form. y=a(xβˆ’h)2+ky = a(x - h)^2 + k. Vertex at (h,k)(h, k).

Vertex from standard form. xv=βˆ’b/(2a)x_v = -b/(2a). yv=cβˆ’b2/(4a)y_v = c - b^2/(4a).

The discriminant

Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac.

IMATH_19 Roots Parabola
IMATH_20 Two real distinct Crosses xx-axis twice
IMATH_22 One repeated Touches xx-axis
IMATH_24 No real Does not touch

Quadratic formula

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

Sketching a parabola

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

Worked example

Find the equation of the line through (1,4)(1, 4) and (3,βˆ’2)(3, -2) in y=mx+cy = mx + c form.

m=(βˆ’2βˆ’4)/(3βˆ’1)=βˆ’6/2=βˆ’3m = (-2 - 4)/(3 - 1) = -6/2 = -3.

Point-slope: yβˆ’4=βˆ’3(xβˆ’1)y - 4 = -3(x - 1). Expand: y=βˆ’3x+7y = -3x + 7.

Common traps

Sign on βˆ’b/(2a)-b/(2a). Common slip is to drop the minus.

Vertex form sign of hh. (xβˆ’3)2(x - 3)^2 has h=3h = 3, not βˆ’3-3.

Confusing aa with leading coefficient. Always same in standard and vertex forms.

Quadratic-formula sign error. The denominator is 2a2a for the whole expression.

In one sentence

Linear functions y=mx+cy = mx + c have gradient mm (parallel lines share mm, perpendicular have m1m2=βˆ’1m_1 m_2 = -1), and quadratics y=ax2+bx+cy = ax^2 + bx + c have vertex at x=βˆ’b/(2a)x = -b/(2a) and discriminant Ξ”=b2βˆ’4ac\Delta = b^2 - 4ac that classifies the roots (>0> 0 two real, =0= 0 one repeated, <0< 0 none).

Past exam questions, worked

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

Year 11 SAC4 marksFor the quadratic $f(x) = 2x^2 - 8x + 5$, 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=βˆ’3y_v = 2(4) - 8(2) + 5 = -3. Vertex: (2,βˆ’3)(2, -3).

(b) Discriminant. Ξ”=b2βˆ’4ac=64βˆ’40=24>0\Delta = b^2 - 4ac = 64 - 40 = 24 > 0, two real distinct roots.

(c) Roots. x=8Β±244=4Β±62=2Β±62x = \dfrac{8 \pm \sqrt{24}}{4} = \dfrac{4 \pm \sqrt 6}{2} = 2 \pm \dfrac{\sqrt 6}{2}.

Markers reward xv=βˆ’b/(2a)x_v = -b/(2a), discriminant, and surd simplification.

Related dot points