ποΈ Practice Workbook β BRAND-NEW problems
Same 4 topics as the exam Β· different numbers she's never seen
Do them on paper (or in the boxes), then open the answer key to check.
Rule: don't peek at the answers until you've fully tried each one. That's how training works.
How to train with this:
- Print it, or write in the grey boxes / on paper.
- Show every step β the real exam gives marks for working, not just the answer.
- Finish a whole topic, then open its β Answer key and mark yourself.
- Anything you got wrong β redo it tomorrow from scratch.
1 Solve these quadratic equations
Move everything to one side first. Then pick: common factor, square root, or the formula. Watch for "no real solution".
axΒ² β 5x = 0
b0 = 3xΒ² + 12x
cxΒ² β 7x + 10 = 0
dxΒ² + 2x β 15 = 0
e(x β 2)Β² = 9
f2(x + 1)Β² = 8
g(x + 4)Β² β 5 = 0
hxΒ² + 4x + 1 = 0
i2xΒ² β 3x β 2 = 0
jxΒ² β 6x + 9 = 0
kxΒ² + 2x + 5 = 0
l(2x β 1)(x + 3) = 0
β Answer key β Topic 1
| # | Answer | How |
|---|---|---|
| a | x = 0, x = 5 | factor x(xβ5) |
| b | x = 0, x = β4 | factor 3x(x+4) |
| c | x = 2, x = 5 | (xβ2)(xβ5) |
| d | x = β5, x = 3 | (x+5)(xβ3) |
| e | x = 5, x = β1 | β: xβ2 = Β±3 |
| f | x = 1, x = β3 | (x+1)Β²=4 β Β±2 |
| g | x = β4 Β± β5 β β1.76, β6.24 | β method |
| h | x = β2 Β± β3 β β0.27, β3.73 | formula, Ξ=12 |
| i | x = 2, x = β0.5 | formula Ξ=25, or (2x+1)(xβ2) |
| j | x = 3 (only one) | (xβ3)Β², Ξ=0 |
| k | No real solution | Ξ = 4β20 = β16 < 0 |
| l | x = 0.5, x = β3 | already factored |
2 Analyze each function fully
For each one find: domain Β· opens up/down Β· zeros Β· vertex Β· y-intercept Β· image Β· positivity/negativity, and draw the graph.
af(x) = (x β 1)(x + 5)
bf(x) = xΒ² β 4x β 5
cf(x) = β(x β 2)Β² + 9
df(x) = xΒ² β 16
ef(x) = 2(x + 1)Β² β 8
ff(x) = 3x(x β 2)
β Answer key β Topic 2
| f(x) | Opens | Zeros | Vertex | Y-int | Image | Positive / Negative |
|---|---|---|---|---|---|---|
| a) (xβ1)(x+5) | up | 1, β5 | (β2, β9) | β5 | y β₯ β9 | + : x<β5 or x>1 β : β5<x<1 |
| b) xΒ²β4xβ5 | up | 5, β1 | (2, β9) | β5 | y β₯ β9 | + : x<β1 or x>5 β : β1<x<5 |
| c) β(xβ2)Β²+9 | down | 5, β1 | (2, 9) | 5 | y β€ 9 | + : β1<x<5 β : x<β1 or x>5 |
| d) xΒ²β16 | up | 4, β4 | (0, β16) | β16 | y β₯ β16 | + : x<β4 or x>4 β : β4<x<4 |
| e) 2(x+1)Β²β8 | up | 1, β3 | (β1, β8) | β6 | y β₯ β8 | + : x<β3 or x>1 β : β3<x<1 |
| f) 3x(xβ2) | up | 0, 2 | (1, β3) | 0 | y β₯ β3 | + : x<0 or x>2 β : 0<x<2 |
Dominio de todas: β. Para graficar: marcΓ‘ raΓces + vΓ©rtice + ordenada al origen y unΓ con curva suave.
π The 6 graphs β compare with your drawings
a) (xβ1)(x+5)
b) xΒ²β4xβ5
c) β(xβ2)Β²+9
d) xΒ²β16
e) 2(x+1)Β²β8
f) 3x(xβ2)
3 Word / graph problems
Spot the pattern: factored form β intercepts & vertex Β· vertex form β find d Β· complete the square / discriminant = 0.
1
The parabola y = (5 β x)(x + 1) has x-intercepts A and C, and maximum point B. Find A, B and C.
2
f(x) = (x β p)(x β q) cuts the x-axis at β3 and 1. (a) Write p and q. (b) Find the x-coordinate of the minimum.
3
y = d(x β m)Β² + p has x-intercepts (2, 0) and (8, 0) and vertex V(m, 4). (a) Write m and p. (b) Find d.
4
Express f(x) = xΒ² β 6x + 11 in the form (x β h)Β² + k, and state the minimum value.
5
f(x) = a(x β 2)Β² β 5 and f(0) = 3. (a) Vertex? (b) Find a.
6
xΒ² + 6x + k = 0 has exactly one solution. Find k.
7
9xΒ² + 6kx + 4 = 0 (with k > 0) has exactly one solution. Find k.
β Answer key β Topic 3
| # | Answer | Key step |
|---|---|---|
| 1 | A(β1, 0), B(2, 9), C(5, 0) | zeros 5 & β1; vertex x = midpoint = 2, y = 3Β·3 = 9 |
| 2 | (a) p = β3, q = 1 (b) x = β1 | p,q are the intercepts; min x = midpoint of β3 & 1 |
| 3 | (a) m = 5, p = 4 (b) d = β4/9 β β0.44 | m = midpoint of 2 & 8; put (2,0): 0 = 9d + 4 |
| 4 | (x β 3)Β² + 2; minimum value = 2 | xΒ²β6x = (xβ3)Β²β9, then β9+11=2 |
| 5 | (a) vertex (2, β5) (b) a = 2 | f(0)=3: a(4)β5=3 β 4a=8 |
| 6 | k = 9 | Ξ = 36 β 4k = 0 |
| 7 | k = 2 | Ξ = 36kΒ² β 144 = 0 β kΒ² = 4, k > 0 |
4 Systems with 3 unknowns
Tidy brackets first β eliminate one variable β solve the 2Γ2 β back-substitute β check. Show all working.
A
x + y + z = 6
x β y + z = 2
2x + y β z = 1
x β y + z = 2
2x + y β z = 1
B
2x + y β z = 3
x β y + 2z = 5
3x + 2y + z = 10
x β y + 2z = 5
3x + 2y + z = 10
C
x + y + z = 6
2(x β y) = z + 2
3x β y + z = 10
2(x β y) = z + 2
3x β y + z = 10
β Answer key β Topic 4
| System | Solution (x, y, z) | Quick check |
|---|---|---|
| A | x = 1, y = 2, z = 3 | 1+2+3 = 6 β |
| B | x = 2, y = 1, z = 2 | 3(2)+2(1)+2 = 10 β |
| C | x = 3, y = 1, z = 2 | 2(3β1)=4 = 2+2 β |
Tip: en C, ordenΓ‘ la 2Βͺ como 2x β 2y β z = 2 antes de empezar.
π Full mock exam (do it timed, like Monday)
Give yourself 60 minutes, no answer key open, calculator + pen. This mixes all 4 topics in exam order. Mark it after with the keys above-style answers below.
1
Solve: (a) xΒ² + 3x β 10 = 0 (b) 3(x β 2)Β² = 12 (c) 2xΒ² + x β 6 = 0
2
Fully analyze f(x) = xΒ² β 2x β 8 (domain, zeros, vertex, y-int, image, positivity/negativity) and graph it.
3
f(x) = a(x β 3)Β² β 2 and f(1) = 6. Find the vertex, find a, and the y-intercept.
4
Solve the system: x + y + z = 4 ; 2x β y + z = 5 ; x + 2y β z = β3
β Answer key β Full mock
| Q | Answer |
|---|---|
| 1a | (x+5)(xβ2) β x = β5, x = 2 |
| 1b | (xβ2)Β² = 4 β x = 4, x = 0 |
| 1c | Ξ = 49 β x = 1.5, x = β2 [(2xβ3)(x+2)] |
| 2 | Domain β Β· opens up Β· zeros 4 & β2 Β· vertex (1, β9) Β· y-int β8 Β· image y β₯ β9 Β· + : x<β2 or x>4, β : β2<x<4 |
| 3 | vertex (3, β2); f(1)=6 β 4aβ2=6 β a = 2; y-int f(0)=2(9)β2 = 16 |
| 4 | x = 1, y = β1, z = 4 (check: 1β1+4=4 β) |
Scoring her mock: count topics, not just final numbers. If she nails 1, 2 and 4 but stumbles on 3, spend extra time on vertex-form problems. The pattern of her mistakes tells you exactly what to drill before Monday.
π Si llegaste hasta acΓ‘, Faustinaβ¦
Ya hiciste lo mΓ‘s difΓcil: practicar. Estoy infinitamente orgulloso de vos. Pase lo que pase el lunes, para mΓ ya ganaste β igual que ganamos esa noche bailando en tus 15. Te amo muchΓsimo. β PapΓ‘ β€οΈπ€β€οΈ
β½ Γltima de PapΓ‘: encarΓ‘ el simulacro como River la final de Madrid 2018 β con huevo y con cabeza. River 3 β Boca 1, y vos: 10 β examen 0. πͺβ€οΈπ€