Matemáticas 2 · Tema 6
Funciones cuadráticas
Mini apunte
Una función cuadrática es de la forma $f(x) = ax^2 + bx + c$. Para graficarla en un intervalo dado, construimos una tabla de valores calculando la imagen (el resultado de evaluar) cada número entero en dicho intervalo. La representación gráfica es una curva llamada **parábola**:
- Si $a > 0$ (coeficiente principal positivo), la parábola abre **hacia arriba** (tiene un punto mínimo).
- Si $a < 0$ (coeficiente principal negativo), la parábola abre **hacia abajo** (tiene un punto máximo).
Ejemplo
Función: $f(x) = x^2 - 6x + 5$
Intervalo: $f: [0, 6] o \mathbb{R}$
| x | f(x) |
|---|---|
| 0 | 5 |
| 1 | 0 |
| 2 | −3 |
| 3 | −4 |
| 4 | −3 |
| 5 | 0 |
| 6 | 5 |
$f(0) = (0)^2 - 6(0) + 5 = $ $5$
$f(1) = (1)^2 - 6(1) + 5 = 1-6+5 = $ $0$
$f(2) = (2)^2 - 6(2) + 5 = 4-12+5 = $ $-3$
$f(3) = (3)^2 - 6(3) + 5 = 9-18+5 = $ $-4$
$f(4) = (4)^2 - 6(4) + 5 = 16-24+5 = $ $-3$
$f(5) = (5)^2 - 6(5) + 5 = 25-30+5 = $ $0$
$f(6) = (6)^2 - 6(6) + 5 = 36-36+5 = $ $5$
$f(1) = (1)^2 - 6(1) + 5 = 1-6+5 = $ $0$
$f(2) = (2)^2 - 6(2) + 5 = 4-12+5 = $ $-3$
$f(3) = (3)^2 - 6(3) + 5 = 9-18+5 = $ $-4$
$f(4) = (4)^2 - 6(4) + 5 = 16-24+5 = $ $-3$
$f(5) = (5)^2 - 6(5) + 5 = 25-30+5 = $ $0$
$f(6) = (6)^2 - 6(6) + 5 = 36-36+5 = $ $5$
Ejercicios
Bloque 1
$f(x) = x^2 - 2x - 8, \quad [-2, 4]$
$f(x) = x^2 - 6x + 8, \quad [0, 6]$
$f(x) = x^2 + 4x, \quad [-5, 1]$
$f(x) = -x^2 - 4x, \quad [-5, 1]$
$f(x) = -x^2 + 2x + 8, \quad [-2, 4]$
Bloque 2
$f(x) = -x^2 - 10x - 24, \quad [-8, -2]$
$f(x) = x^2 - 4x + 4, \quad [-1, 5]$
$f(x) = x^2 + 6x + 9, \quad [-6, 0]$
$f(x) = -x^2 + 4x - 3, \quad [-1, 5]$
$f(x) = x^2 - 7x + 10, \quad [1, 6]$
Resoluciones
▼Bloque 1
$f(x) = x^2 - 2x - 8$ en $[-2, 4]$
| x | f(x) |
|---|---|
| −2 | 0 |
| −1 | −5 |
| 0 | −8 |
| 1 | −9 |
| 2 | −8 |
| 3 | −5 |
| 4 | 0 |
$f(-2) = (-2)^2 - 2(-2) - 8 = 4+4-8 = $ $0$
$f(-1) = (-1)^2 - 2(-1) - 8 = 1+2-8 = $ $-5$
$f(0) = (0)^2 - 2(0) - 8 = $ $-8$
$f(1) = (1)^2 - 2(1) - 8 = 1-2-8 = $ $-9$
$f(2) = (2)^2 - 2(2) - 8 = 4-4-8 = $ $-8$
$f(3) = (3)^2 - 2(3) - 8 = 9-6-8 = $ $-5$
$f(4) = (4)^2 - 2(4) - 8 = 16-8-8 = $ $0$
$f(-1) = (-1)^2 - 2(-1) - 8 = 1+2-8 = $ $-5$
$f(0) = (0)^2 - 2(0) - 8 = $ $-8$
$f(1) = (1)^2 - 2(1) - 8 = 1-2-8 = $ $-9$
$f(2) = (2)^2 - 2(2) - 8 = 4-4-8 = $ $-8$
$f(3) = (3)^2 - 2(3) - 8 = 9-6-8 = $ $-5$
$f(4) = (4)^2 - 2(4) - 8 = 16-8-8 = $ $0$
$f(x) = x^2 - 6x + 8$ en $[0, 6]$
| x | f(x) |
|---|---|
| 0 | 8 |
| 1 | 3 |
| 2 | 0 |
| 3 | −1 |
| 4 | 0 |
| 5 | 3 |
| 6 | 8 |
$f(0) = (0)^2 - 6(0) + 8 = $ $8$
$f(1) = (1)^2 - 6(1) + 8 = 1-6+8 = $ $3$
$f(2) = (2)^2 - 6(2) + 8 = 4-12+8 = $ $0$
$f(3) = (3)^2 - 6(3) + 8 = 9-18+8 = $ $-1$
$f(4) = (4)^2 - 6(4) + 8 = 16-24+8 = $ $0$
$f(5) = (5)^2 - 6(5) + 8 = 25-30+8 = $ $3$
$f(6) = (6)^2 - 6(6) + 8 = 36-36+8 = $ $8$
$f(1) = (1)^2 - 6(1) + 8 = 1-6+8 = $ $3$
$f(2) = (2)^2 - 6(2) + 8 = 4-12+8 = $ $0$
$f(3) = (3)^2 - 6(3) + 8 = 9-18+8 = $ $-1$
$f(4) = (4)^2 - 6(4) + 8 = 16-24+8 = $ $0$
$f(5) = (5)^2 - 6(5) + 8 = 25-30+8 = $ $3$
$f(6) = (6)^2 - 6(6) + 8 = 36-36+8 = $ $8$
$f(x) = x^2 + 4x$ en $[-5, 1]$
| x | f(x) |
|---|---|
| −5 | 5 |
| −4 | 0 |
| −3 | −3 |
| −2 | −4 |
| −1 | −3 |
| 0 | 0 |
| 1 | 5 |
$f(-5) = (-5)^2 + 4(-5) = 25-20 = $ $5$
$f(-4) = (-4)^2 + 4(-4) = 16-16 = $ $0$
$f(-3) = (-3)^2 + 4(-3) = 9-12 = $ $-3$
$f(-2) = (-2)^2 + 4(-2) = 4-8 = $ $-4$
$f(-1) = (-1)^2 + 4(-1) = 1-4 = $ $-3$
$f(0) = (0)^2 + 4(0) = $ $0$
$f(1) = (1)^2 + 4(1) = 1+4 = $ $5$
$f(-4) = (-4)^2 + 4(-4) = 16-16 = $ $0$
$f(-3) = (-3)^2 + 4(-3) = 9-12 = $ $-3$
$f(-2) = (-2)^2 + 4(-2) = 4-8 = $ $-4$
$f(-1) = (-1)^2 + 4(-1) = 1-4 = $ $-3$
$f(0) = (0)^2 + 4(0) = $ $0$
$f(1) = (1)^2 + 4(1) = 1+4 = $ $5$
$f(x) = -x^2 - 4x$ en $[-5, 1]$
| x | f(x) |
|---|---|
| −5 | −5 |
| −4 | 0 |
| −3 | 3 |
| −2 | 4 |
| −1 | 3 |
| 0 | 0 |
| 1 | −5 |
$f(-5) = -(-5)^2 - 4(-5) = -25+20 = $ $-5$
$f(-4) = -(-4)^2 - 4(-4) = -16+16 = $ $0$
$f(-3) = -(-3)^2 - 4(-3) = -9+12 = $ $3$
$f(-2) = -(-2)^2 - 4(-2) = -4+8 = $ $4$
$f(-1) = -(-1)^2 - 4(-1) = -1+4 = $ $3$
$f(0) = -(0)^2 - 4(0) = $ $0$
$f(1) = -(1)^2 - 4(1) = -1-4 = $ $-5$
$f(-4) = -(-4)^2 - 4(-4) = -16+16 = $ $0$
$f(-3) = -(-3)^2 - 4(-3) = -9+12 = $ $3$
$f(-2) = -(-2)^2 - 4(-2) = -4+8 = $ $4$
$f(-1) = -(-1)^2 - 4(-1) = -1+4 = $ $3$
$f(0) = -(0)^2 - 4(0) = $ $0$
$f(1) = -(1)^2 - 4(1) = -1-4 = $ $-5$
$f(x) = -x^2 + 2x + 8$ en $[-2, 4]$
| x | f(x) |
|---|---|
| −2 | 0 |
| −1 | 5 |
| 0 | 8 |
| 1 | 9 |
| 2 | 8 |
| 3 | 5 |
| 4 | 0 |
$f(-2) = -(-2)^2 + 2(-2) + 8 = -4-4+8 = $ $0$
$f(-1) = -(-1)^2 + 2(-1) + 8 = -1-2+8 = $ $5$
$f(0) = -(0)^2 + 2(0) + 8 = $ $8$
$f(1) = -(1)^2 + 2(1) + 8 = -1+2+8 = $ $9$
$f(2) = -(2)^2 + 2(2) + 8 = -4+4+8 = $ $8$
$f(3) = -(3)^2 + 2(3) + 8 = -9+6+8 = $ $5$
$f(4) = -(4)^2 + 2(4) + 8 = -16+8+8 = $ $0$
$f(-1) = -(-1)^2 + 2(-1) + 8 = -1-2+8 = $ $5$
$f(0) = -(0)^2 + 2(0) + 8 = $ $8$
$f(1) = -(1)^2 + 2(1) + 8 = -1+2+8 = $ $9$
$f(2) = -(2)^2 + 2(2) + 8 = -4+4+8 = $ $8$
$f(3) = -(3)^2 + 2(3) + 8 = -9+6+8 = $ $5$
$f(4) = -(4)^2 + 2(4) + 8 = -16+8+8 = $ $0$
Bloque 2
$f(x) = -x^2 - 10x - 24$ en $[-8, -2]$
| x | f(x) |
|---|---|
| −8 | −8 |
| −7 | −3 |
| −6 | 0 |
| −5 | 1 |
| −4 | 0 |
| −3 | −3 |
| −2 | −8 |
$f(-8) = -(-8)^2 - 10(-8) - 24 = -64+80-24 = $ $-8$
$f(-7) = -(-7)^2 - 10(-7) - 24 = -49+70-24 = $ $-3$
$f(-6) = -(-6)^2 - 10(-6) - 24 = -36+60-24 = $ $0$
$f(-5) = -(-5)^2 - 10(-5) - 24 = -25+50-24 = $ $1$
$f(-4) = -(-4)^2 - 10(-4) - 24 = -16+40-24 = $ $0$
$f(-3) = -(-3)^2 - 10(-3) - 24 = -9+30-24 = $ $-3$
$f(-2) = -(-2)^2 - 10(-2) - 24 = -4+20-24 = $ $-8$
$f(-7) = -(-7)^2 - 10(-7) - 24 = -49+70-24 = $ $-3$
$f(-6) = -(-6)^2 - 10(-6) - 24 = -36+60-24 = $ $0$
$f(-5) = -(-5)^2 - 10(-5) - 24 = -25+50-24 = $ $1$
$f(-4) = -(-4)^2 - 10(-4) - 24 = -16+40-24 = $ $0$
$f(-3) = -(-3)^2 - 10(-3) - 24 = -9+30-24 = $ $-3$
$f(-2) = -(-2)^2 - 10(-2) - 24 = -4+20-24 = $ $-8$
$f(x) = x^2 - 4x + 4$ en $[-1, 5]$
| x | f(x) |
|---|---|
| −1 | 9 |
| 0 | 4 |
| 1 | 1 |
| 2 | 0 |
| 3 | 1 |
| 4 | 4 |
| 5 | 9 |
$f(-1) = (-1)^2 - 4(-1) + 4 = 1+4+4 = $ $9$
$f(0) = (0)^2 - 4(0) + 4 = $ $4$
$f(1) = (1)^2 - 4(1) + 4 = 1-4+4 = $ $1$
$f(2) = (2)^2 - 4(2) + 4 = 4-8+4 = $ $0$
$f(3) = (3)^2 - 4(3) + 4 = 9-12+4 = $ $1$
$f(4) = (4)^2 - 4(4) + 4 = 16-16+4 = $ $4$
$f(5) = (5)^2 - 4(5) + 4 = 25-20+4 = $ $9$
$f(0) = (0)^2 - 4(0) + 4 = $ $4$
$f(1) = (1)^2 - 4(1) + 4 = 1-4+4 = $ $1$
$f(2) = (2)^2 - 4(2) + 4 = 4-8+4 = $ $0$
$f(3) = (3)^2 - 4(3) + 4 = 9-12+4 = $ $1$
$f(4) = (4)^2 - 4(4) + 4 = 16-16+4 = $ $4$
$f(5) = (5)^2 - 4(5) + 4 = 25-20+4 = $ $9$
$f(x) = x^2 + 6x + 9$ en $[-6, 0]$
| x | f(x) |
|---|---|
| −6 | 9 |
| −5 | 4 |
| −4 | 1 |
| −3 | 0 |
| −2 | 1 |
| −1 | 4 |
| 0 | 9 |
$f(-6) = (-6)^2 + 6(-6) + 9 = 36-36+9 = $ $9$
$f(-5) = (-5)^2 + 6(-5) + 9 = 25-30+9 = $ $4$
$f(-4) = (-4)^2 + 6(-4) + 9 = 16-24+9 = $ $1$
$f(-3) = (-3)^2 + 6(-3) + 9 = 9-18+9 = $ $0$
$f(-2) = (-2)^2 + 6(-2) + 9 = 4-12+9 = $ $1$
$f(-1) = (-1)^2 + 6(-1) + 9 = 1-6+9 = $ $4$
$f(0) = (0)^2 + 6(0) + 9 = $ $9$
$f(-5) = (-5)^2 + 6(-5) + 9 = 25-30+9 = $ $4$
$f(-4) = (-4)^2 + 6(-4) + 9 = 16-24+9 = $ $1$
$f(-3) = (-3)^2 + 6(-3) + 9 = 9-18+9 = $ $0$
$f(-2) = (-2)^2 + 6(-2) + 9 = 4-12+9 = $ $1$
$f(-1) = (-1)^2 + 6(-1) + 9 = 1-6+9 = $ $4$
$f(0) = (0)^2 + 6(0) + 9 = $ $9$
$f(x) = -x^2 + 4x - 3$ en $[-1, 5]$
| x | f(x) |
|---|---|
| −1 | −8 |
| 0 | −3 |
| 1 | 0 |
| 2 | 1 |
| 3 | 0 |
| 4 | −3 |
| 5 | −8 |
$f(-1) = -(-1)^2 + 4(-1) - 3 = -1-4-3 = $ $-8$
$f(0) = -(0)^2 + 4(0) - 3 = $ $-3$
$f(1) = -(1)^2 + 4(1) - 3 = -1+4-3 = $ $0$
$f(2) = -(2)^2 + 4(2) - 3 = -4+8-3 = $ $1$
$f(3) = -(3)^2 + 4(3) - 3 = -9+12-3 = $ $0$
$f(4) = -(4)^2 + 4(4) - 3 = -16+16-3 = $ $-3$
$f(5) = -(5)^2 + 4(5) - 3 = -25+20-3 = $ $-8$
$f(0) = -(0)^2 + 4(0) - 3 = $ $-3$
$f(1) = -(1)^2 + 4(1) - 3 = -1+4-3 = $ $0$
$f(2) = -(2)^2 + 4(2) - 3 = -4+8-3 = $ $1$
$f(3) = -(3)^2 + 4(3) - 3 = -9+12-3 = $ $0$
$f(4) = -(4)^2 + 4(4) - 3 = -16+16-3 = $ $-3$
$f(5) = -(5)^2 + 4(5) - 3 = -25+20-3 = $ $-8$
$f(x) = x^2 - 7x + 10$ en $[1, 6]$
| x | f(x) |
|---|---|
| 1 | 4 |
| 2 | 0 |
| 3 | −2 |
| 4 | −2 |
| 5 | 0 |
| 6 | 4 |
$f(1) = (1)^2 - 7(1) + 10 = 1-7+10 = $ $4$
$f(2) = (2)^2 - 7(2) + 10 = 4-14+10 = $ $0$
$f(3) = (3)^2 - 7(3) + 10 = 9-21+10 = $ $-2$
$f(4) = (4)^2 - 7(4) + 10 = 16-28+10 = $ $-2$
$f(5) = (5)^2 - 7(5) + 10 = 25-35+10 = $ $0$
$f(6) = (6)^2 - 7(6) + 10 = 36-42+10 = $ $4$
$f(2) = (2)^2 - 7(2) + 10 = 4-14+10 = $ $0$
$f(3) = (3)^2 - 7(3) + 10 = 9-21+10 = $ $-2$
$f(4) = (4)^2 - 7(4) + 10 = 16-28+10 = $ $-2$
$f(5) = (5)^2 - 7(5) + 10 = 25-35+10 = $ $0$
$f(6) = (6)^2 - 7(6) + 10 = 36-42+10 = $ $4$
Ejercicios extra
▼Bloque 1
$f(x) = x^2 - x - 2, \quad [-2, 3]$
$f(x) = -x^2 + 9, \quad [-4, 4]$
$f(x) = x^2 - 4x + 3, \quad [0, 4]$
Bloque 2
$f(x) = 2x^2 - 4x, \quad [-1, 3]$
$f(x) = -x^2 - 2x + 3, \quad [-4, 2]$
$f(x) = x^2 + 2x + 1, \quad [-3, 1]$
Bloque 3
$f(x) = -2x^2 + 8, \quad [-3, 3]$
$f(x) = x^2 - 9, \quad [-4, 4]$
$f(x) = -x^2 + 6x - 8, \quad [1, 5]$
Bloque 4
$f(x) = x^2 + 5x + 6, \quad [-5, 0]$
$f(x) = -x^2 + x + 6, \quad [-3, 4]$
$f(x) = x^2 - 2x + 1, \quad [-1, 3]$