🎯 Math Exam Survival Guide

Practice n°3 — 3º NES · Colegio Jesús María

Topic: Quadratic equations, quadratic functions & systems of equations

What's on the test: everything up to Exercise 4 (Ex. 5 & 6 are NOT on it).

You've got this. Read each rule → copy the worked example → do the practice → check the answers.

How to use this guide (read this first)

4 topics only. Master these four and you cover the whole test.
Each topic has: a RULE box (the recipe), a WORKED example (copy the steps), and ANSWERS to every real exercise so you can check yourself.
Always show your working. The sheet literally says "All working must be shown." Half the marks are for the steps, not just the final number.
Las palabras en violeta son la traducción de los términos clave.

1 · Solve equations 2 · Functions 3 · Word problems 4 · 3-variable systems ⭐ Cheat sheet 📝 Mock test

1 Solving quadratic equations

Resolver ecuaciones cuadráticas (las que tienen x²).

STEP 0 — Always first: move everything to one side so it looks like ax² + bx + c = 0 (one side = 0).

You have 3 tools. Pick by what the equation looks like:

Tool A — Common factor (when there's NO lone number, c = 0)

x² + 2x = 0 → x(x + 2) = 0 → x = 0 or x + 2 = 0 → x = 0, x = −2

Sacás factor común x. Una solución casi siempre es x = 0.

Tool B — Square-root method (when it looks like ( )² = number)

2(x−1)² = 3 → (x−1)² = 1.5 → x−1 = ±√1.5 → x = 1 ± 1.22

⚠️ Don't forget the ± (plus AND minus). Forgetting it loses you a whole solution. If the number under the root is negative → no real solution.

Tool C — The Quadratic Formula (always works, use when A and B don't fit)

x = ( −b ± √(b² − 4ac) ) / (2a)

First write down a, b, c clearly.
The part under the root, b²−4ac, is the discriminant (Δ):
Δ > 0 → two solutions · Δ = 0 → one solution · Δ < 0 → no real solution.

Si ya está factoreado, como (2x−4)(x+1)=0, igualás cada paréntesis a 0.

✍️ Fully worked example — equation 1b: −3 = 1 − x(x+1)

Expand: −3 = 1 − x² − x

Move all to one side: x² + x − 1 − 3 = 0 → x² + x − 4 = 0

a = 1, b = 1, c = −4. Δ = 1² − 4(1)(−4) = 1 + 16 = 17 (positive → 2 answers)

x = (−1 ± √17) / 2 = (−1 ± 4.12) / 2

x ≈ 1.56 or x ≈ −2.56

✅ ANSWERS — Exercise 1 (a–i). Try each first, THEN open.

#	Equation	Best tool	Answers
a	0 = 2x + x²	A (factor)	x = 0, x = −2
b	−3 = 1 − x(x+1)	C (formula)	x ≈ 1.56, x ≈ −2.56
c	−6 = x² + 5x	C or factor	x = −2, x = −3
d	3 = 2(x−1)²	B (√)	x = 1 ± 1.22 → ≈ 2.22, −0.22
e	4 = −3x(x+2)	C	Δ = −12 < 0 → no real solution
f	0 = (x−3)² − 2	B (√)	x = 3 ± √2 → ≈ 4.41, 1.59
g	0 = (2x−4)(x+1)	already factored	x = 2, x = −1
h	3 = 4(x−5)²	B (√)	x = 5 ± 0.87 → ≈ 5.87, 4.13
i	0 = 3x² + 2x − 4	C	x = (−1 ± √13)/3 → ≈ 0.87, −1.54

2 Analyzing quadratic functions

Estudiar la parábola: dominio, imagen, vértice, raíces, ordenada al origen, gráfico, positividad y negatividad.

For EVERY function they give you, you must find the same 7 things. Here is the exact checklist and how to get each one.

What to find	How to get it
Domain Dominio	Always all real numbers (ℝ) for a parabola. Easy mark.
Opens up or down?	Look at a (number in front of x²). a > 0 → opens up (U, minimum). a < 0 → opens down (∩, maximum).
Zeros / roots Raíces / ceros	Set f(x) = 0 and solve (Topic 1!). These are where it crosses the x-axis.
Vertex Vértice	x of vertex = −b/(2a) (or the midpoint between the two zeros). Then plug that x back in to get y. Write it as a point (x, y).
Y-intercept Ordenada al origen	Just calculate f(0) (it's the value of c).
Image / Range Imagen	Use the vertex y-value. Opens up → y ≥ y_vertex. Opens down → y ≤ y_vertex.
Positivity / Negativity Positividad / Negatividad	Where is the graph above (+) or below (−) the x-axis? Use the zeros as borders + the up/down shape. (See rule below.)

Positivity/Negativity shortcut (parabola with 2 zeros r₁ < r₂):
• Opens UP: positive outside the zeros (x < r₁ or x > r₂), negative between them.
• Opens DOWN: positive between the zeros, negative outside.
Trick: a quick sketch makes this obvious — that's why we graph it.

✍️ Fully worked example — function 2a: f(x) = (x−3)(x+4)

Domain: ℝ (all reals)

Opens: if you expand you get x²+x−12, a = 1 > 0 → opens UP (minimum)

Zeros: (x−3)(x+4)=0 → x = 3 and x = −4

Vertex x: midpoint of zeros = (3 + (−4))/2 = −0.5. Vertex y: f(−0.5) = (−3.5)(3.5) = −12.25 → Vertex (−0.5, −12.25)

Y-intercept: f(0) = (−3)(4) = −12 → point (0, −12)

Image: opens up → y ≥ −12.25

Positivity: x < −4 or x > 3 · Negativity: −4 < x < 3

📈 Graph: plot the two zeros (−4,0) and (3,0), the vertex (−0.5,−12.25), and the y-intercept (0,−12). Join with a smooth U-shape — exactly like this:

✅ ANSWERS — Exercise 2 (a–f). Full table.

f(x)	Opens	Zeros	Vertex	Y-int	Image	Positive / Negative
a) (x−3)(x+4)	up	−4, 3	(−0.5, −12.25)	−12	y ≥ −12.25	+ : x<−4 or x>3 − : −4<x<3
b) 2x²+2x−24	up	−4, 3	(−0.5, −24.5)	−24	y ≥ −24.5	+ : x<−4 or x>3 − : −4<x<3
c) x²−9	up	−3, 3	(0, −9)	−9	y ≥ −9	+ : x<−3 or x>3 − : −3<x<3
d) 2−(x+3)²	down	−3±√2 ≈ −4.41, −1.59	(−3, 2)	−7	y ≤ 2	+ : −4.41<x<−1.59 − : outside that
e) 2x(x−4)	up	0, 4	(2, −8)	0	y ≥ −8	+ : x<0 or x>4 − : 0<x<4
f) 2(x−3)²−4	up	3±√2 ≈ 1.59, 4.41	(3, −4)	14	y ≥ −4	+ : x<1.59 or x>4.41 − : 1.59<x<4.41

Recordá: en d y f ya te dan el vértice "gratis" porque están en forma a(x−h)²+k → vértice (h, k).

📈 The 6 graphs — check your drawings against these

a) (x−3)(x+4)

b) 2x²+2x−24

c) x²−9

d) 2−(x+3)²

e) 2x(x−4)

f) 2(x−3)²−4

Reading the two function forms instantly:
• Factored form a(x−r₁)(x−r₂) → the zeros r₁, r₂ are right there.
• Vertex form a(x−h)²+k → the vertex (h, k) is right there. (¡Ojo con los signos! (x−3) → h=+3)

3 Word / graph problems (Problems 1–7)

Problemas con parábolas. Son siempre los mismos 3 trucos. Identificá cuál te piden.

The 3 patterns that cover all 7 problems

Pattern 1 — Find intercepts & vertex from factored form y=(x−a)(x−b): zeros = a and b; vertex x = midpoint of zeros; vertex y = plug it in. (Problems 1, 2, 3)

Pattern 2 — Find unknowns in vertex form y=d(x−m)²+p: m & p come straight from the vertex; find d by plugging in any other known point and solving. (Problems 4, 6)

Pattern 3 — "Complete the square" / "exactly one solution": rewrite ax²+bx+c into vertex form (Problem 5), or set the discriminant Δ = 0 for one solution (Problem 7).

✍️ Fully worked example — Problem 1: y = (7−x)(1+x)

x-intercepts (A and C): 7−x=0 → x=7; 1+x=0 → x=−1 → A(−1, 0), C(7, 0)

B is the maximum (vertex): x = midpoint = (−1+7)/2 = 3

y = (7−3)(1+3) = 4·4 = 16 → B(3, 16)

✅ ANSWERS — Problems 1–7 (with the key step)

Problem	Answer	Key step
1 y=(7−x)(1+x)	A(−1,0), B(3,16), C(7,0)	zeros from factors; vertex x = midpoint
2 f=(x−p)(x−q), cuts at −½ and 2	(a) p=−½, q=2 (b) x of C = ¾ = 0.75	p,q are the x-intercepts; min x = midpoint
3 f=p(x−q)(x−r); thru (−2,0),(0,−4),(4,0)	(a) q,r = −2 and 4 (b) x = 1 (c) p = 0.5	(c) put (0,−4) in: −4 = p(2)(−4) → p=0.5
4 y=d(x−m)²+p; cuts (1,0),(5,0), V(m,2)	(a) m=3, p=2 (b) d = −0.5	m = midpoint of 1&5; put (1,0) in to get d
5 f=2x²−8x+5 (complete square)	(a) 2(x−2)²−3 (b) min = −3	2(x²−4x) → 2(x−2)²−8+5
6 f=a(x−4)²+8, f(7)=−10	(a) vertex (4,8) (b) a=−2 (c) y-int = −24	9a+8=−10 → a=−2; then f(0)
7 4x²+4kx+9=0, one solution, k>0	k = 3	Δ=0: (4k)²−4·4·9=0 → 16k²=144

4 Systems of equations — 3 unknowns

Sistemas de 3 ecuaciones con 3 incógnitas (x, y, z). Hay que mostrar TODO el procedimiento.

The reliable method (substitution / elimination)

Tidy each equation first: multiply out brackets, put it as x + y + z = number.
Eliminate one variable. Easiest: if one equation already gives a variable alone (like z = … or two equations are almost equal), use it. Otherwise add/subtract two equations to cancel one letter.
Now you have 2 equations with 2 unknowns → solve those.
Back-substitute to get the third variable.
CHECK: put x, y, z into the original three equations. (Siempre verificá — es un punto gratis.)

✍️ Fully worked example — System 4a

Tidy them up:
① 2(x+y)=3z−7 → 2x + 2y − 3z = −7
② x − 2y + 2z = 10
③ 3(x−2y+z−4)=9 → x−2y+z−4=3 → x − 2y + z = 7

② − ③: (2z − z) = 10 − 7 → z = 3

Put z=3 in ③: x − 2y + 3 = 7 → x − 2y = 4 → x = 4 + 2y

Put both in ①: 2(4+2y) + 2y − 9 = −7 → 8 + 6y − 9 = −7 → 6y = −6 → y = −1

Then x = 4 + 2(−1) = 2. ✅ Check in ②: 2 −2(−1) +2(3) = 2+2+6 = 10 ✓

Solution: x = 2, y = −1, z = 3

✅ ANSWERS — Exercise 4 (a–d)

System	Solution (x, y, z)
a	x = 2, y = −1, z = 3
b	x = 2, y = −1, z = 3
c	Set up the same way (use −x+2y=12 from the 3rd eq.). ⚠️ Double-check your copy of system c against the sheet — as written it does not give whole numbers, so the equation may be slightly different on paper.
d	x = 4, y = 3, z = −2

Truco: en "a" y "b" la respuesta es la misma (2, −1, 3). Buen control de que vas bien.

⚽ Dato de Papá (River ❤️🤍): resolver un sistema es como River jugando de memoria — vas combinando las piezas hasta que todo encaja y aparece el resultado. Boca todavía está despejando las incógnitas 😅. Vos dale, que te sale.

⭐ One-page cheat sheet (memorize this)

Quadratic formula

x = (−b ± √(b²−4ac)) / 2a

Discriminant Δ = b²−4ac
Δ>0 → 2 sol · Δ=0 → 1 sol · Δ<0 → none

Vertex

x_v = −b/(2a)

then y_v = f(x_v)

Or x_v = midpoint of the two zeros.

The two forms

a(x−r₁)(x−r₂)

→ zeros = r₁, r₂

a(x−h)²+k

→ vertex = (h, k)

Always true for a parabola

Domain = ℝ
Y-intercept = f(0)
a>0 opens up (min) · a<0 opens down (max)
Image: up → y≥k · down → y≤k

Top 5 mistakes that cost marks — don't do these:

Forgetting the ± in the square-root method (you lose a solution).
Sign errors with brackets: (x−3) means the zero/vertex is at +3, not −3.
Not moving everything to one side before using the formula.
Giving the vertex as one number — it's a point (x, y).
Not showing working / not checking the system answer.

📝 Mini mock test (use the "act j" sheet too)

The second sheet, "Activity to hand in – 3J", is the same two topics. Do it as a timed rehearsal, then check below.

✅ ANSWERS — "act j" sheet

1. Solve

a) 3x − x² = 2x²	x = 0, x = 1
b) 2(x−2)² + 3 = 11	x = 4, x = 0
c) x² + 2(x−2) − 1 = 7 + x	x = −4, x = 3

2. Functions

	Opens	Zeros	Vertex	Y-int	Image	Positive / Negative
f(x)=−3x²+3x+6	down	−1, 2	(0.5, 6.75)	6	y ≤ 6.75	+ : −1<x<2 − : x<−1 or x>2
g(x)=2x²−4x−16	up	−2, 4	(1, −18)	−16	y ≥ −18	+ : x<−2 or x>4 − : −2<x<4

Dominio de las dos: ℝ. Graficá usando vértice + raíces + ordenada al origen.

Study plan for the days left (≈30 min/day):
Day 1: Topic 1 — solve all of Ex.1 yourself, then check.
Day 2: Topic 2 — fully analyze 2a, 2c, 2e (+ graph).
Day 3: Topic 3 — do Problems 1, 4, 5, 7 (one of each pattern).
Day 4: Topic 4 — do systems a, b, d completely.
Day 5 (Sunday): Do the whole "act j" sheet timed, like the real exam. Then re-read the cheat sheet before bed.
Monday: Quick cheat-sheet glance. Walk in confident. 💪