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Lic. Lucila Ccapa Maldonado
Función Lineal
Es la función de la forma y = ax + b donde a y b son
constantes a ≠ 0, su grafica de esta función es una
recta.
Gráfica de una función lineal
Se puede hacer de 2 maneras:
1.- Por tabulación
2.- Por método abreviado
Hallar la función por tabulación
Dar valores a la variable “x” y reemplazar en la función dada:
Ejemplo F(x)= 2x -1 o también puede ser y=2x-1
X Y
Pares
ordenados
0 -1 ( 0,-1)
1 1 ( 1, 1 )
2 3 (2,3 )
-1 -3 (-1,-3)
-2 -5 (-2,-5)
Para x=0 remplazo en la
función dada:
y= 2x -1
y=2x-1
y=2(0)-1
y=0-1
y=-1
y así sucesivamente vas
reemplazando con
cualquier número
la función lineal.
Hallar la función por el método abreviado
y= 2x-1
Donde: “2x” es la pendiente y “-1” es el corte en el eje de coordenadas “y”
y=mx+b
1. Hallamos el corte con el eje “y” donde b es igual a “-1”.
2. Ubico en el plano cartesiano -1 que corresponde al eje “y”
3. Hallamos la pendiente “m” como fracción:
𝑚 =
+2
+1
=
𝑦2−𝑦1
𝑥2−𝑥1
=
∆𝑦
∆𝑥
4. Desde el punto de corte me indica lo
siguiente:
Donde “+2” me indica subir 2 hacia
arriba del eje de las coordenadas “
y”.
Donde “+1” me indica recorrer un
espacio hacia la derecha del eje de las
abscisas “x”, entonces la respuesta del
punto es (1,1) con estos 2 puntos ya
podemos graficar la recta.
Hallar la función por tabulación
Dar cualquier valor a “x” y reemplazar en la
función dada.
Ejemplo: x2 -6x +5
X Y Pares ordenados
1 0 (1;0 )
0 5 (0;5 )
2 -3 (2;-3)
-1 12 (-1;12)
-2 21 (-2;21)
Para x=-1 reemplazo
en la función dada:
y= x2–6x+5
y=(-1)2 -6(-1) +5
y= 1+6+5
y=12
Así vas reemplazando
con los otros valores
que has dado y la
respuesta de” y” te va
a salir.
Hallando el vértice y los puntos de corte de la función
es por método abreviado
Vértice: Aplicando la siguiente fórmula: −𝑏
2𝑎
;
−𝑏 + 4𝑎𝑐
4𝑎
Ejemplo: y= x2 -6x +5, donde x2=a,-6x =b,+5=c,
X Y
Vértice de la variable “x” remplazamos en: −𝑏
2𝑎
𝑥𝑣 =
−(−6
2(1
=
6
2
= 3
Vértice de la variable “y” remplazamos en: −𝑏 + 4𝑎𝑐
4𝑎
4
4
4
16
4
20
36
)
1
(
4
)
5
)(
1
(
4
)
6
( 2













v
v
y
y
Entonces el vértice seria (3 ;-4),
ubicamos en el plano cartesiano
Puntos de corte
Ejemplo: y= x2 -6x +5
1) Eje “x” cuando “y” es igual a cero.
0 = x2 -6x +5, aplicando la siguiente fórmula: x=
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
x=
− −6 ± 62−4.1.5
2.1
=
6± 36−20
2
=
6± 16
2
=
6±4
2
Donde “x” tiene 2 soluciones porque tiene 2 signos:
𝑥1 =
6 + 4
2
=
10
2
= 5 𝑥2 =
6 − 4
2
=
2
2
= 1
Puntos de corte: (5;0) y (1;0)
2) Eje “y” cuando “x” es igual a cero.
y = x2 -6x +5
Punto de corte: (0;5)
y = 02 -6(0) +5
y = 0 - 0 +5
y = +5
Puntos de corte:
• “x” cuando “y” es igual
a cero: (5;0) y (1;0)
• “y” cuando “x” es igual
a cero: (0;5)
Ahora graficamos!
Vértice (3 ;-4)
Funcion lineal y cuadratica

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Funcion lineal y cuadratica

  • 1. Lic. Lucila Ccapa Maldonado
  • 2. Función Lineal Es la función de la forma y = ax + b donde a y b son constantes a ≠ 0, su grafica de esta función es una recta. Gráfica de una función lineal Se puede hacer de 2 maneras: 1.- Por tabulación 2.- Por método abreviado
  • 3. Hallar la función por tabulación Dar valores a la variable “x” y reemplazar en la función dada: Ejemplo F(x)= 2x -1 o también puede ser y=2x-1 X Y Pares ordenados 0 -1 ( 0,-1) 1 1 ( 1, 1 ) 2 3 (2,3 ) -1 -3 (-1,-3) -2 -5 (-2,-5) Para x=0 remplazo en la función dada: y= 2x -1 y=2x-1 y=2(0)-1 y=0-1 y=-1 y así sucesivamente vas reemplazando con cualquier número la función lineal.
  • 4. Hallar la función por el método abreviado y= 2x-1 Donde: “2x” es la pendiente y “-1” es el corte en el eje de coordenadas “y” y=mx+b 1. Hallamos el corte con el eje “y” donde b es igual a “-1”. 2. Ubico en el plano cartesiano -1 que corresponde al eje “y” 3. Hallamos la pendiente “m” como fracción: 𝑚 = +2 +1 = 𝑦2−𝑦1 𝑥2−𝑥1 = ∆𝑦 ∆𝑥
  • 5. 4. Desde el punto de corte me indica lo siguiente: Donde “+2” me indica subir 2 hacia arriba del eje de las coordenadas “ y”. Donde “+1” me indica recorrer un espacio hacia la derecha del eje de las abscisas “x”, entonces la respuesta del punto es (1,1) con estos 2 puntos ya podemos graficar la recta.
  • 6.
  • 7. Hallar la función por tabulación Dar cualquier valor a “x” y reemplazar en la función dada. Ejemplo: x2 -6x +5 X Y Pares ordenados 1 0 (1;0 ) 0 5 (0;5 ) 2 -3 (2;-3) -1 12 (-1;12) -2 21 (-2;21) Para x=-1 reemplazo en la función dada: y= x2–6x+5 y=(-1)2 -6(-1) +5 y= 1+6+5 y=12 Así vas reemplazando con los otros valores que has dado y la respuesta de” y” te va a salir.
  • 8. Hallando el vértice y los puntos de corte de la función es por método abreviado Vértice: Aplicando la siguiente fórmula: −𝑏 2𝑎 ; −𝑏 + 4𝑎𝑐 4𝑎 Ejemplo: y= x2 -6x +5, donde x2=a,-6x =b,+5=c, X Y Vértice de la variable “x” remplazamos en: −𝑏 2𝑎 𝑥𝑣 = −(−6 2(1 = 6 2 = 3
  • 9. Vértice de la variable “y” remplazamos en: −𝑏 + 4𝑎𝑐 4𝑎 4 4 4 16 4 20 36 ) 1 ( 4 ) 5 )( 1 ( 4 ) 6 ( 2              v v y y Entonces el vértice seria (3 ;-4), ubicamos en el plano cartesiano
  • 10. Puntos de corte Ejemplo: y= x2 -6x +5 1) Eje “x” cuando “y” es igual a cero. 0 = x2 -6x +5, aplicando la siguiente fórmula: x= −𝑏± 𝑏2−4𝑎𝑐 2𝑎 x= − −6 ± 62−4.1.5 2.1 = 6± 36−20 2 = 6± 16 2 = 6±4 2 Donde “x” tiene 2 soluciones porque tiene 2 signos: 𝑥1 = 6 + 4 2 = 10 2 = 5 𝑥2 = 6 − 4 2 = 2 2 = 1 Puntos de corte: (5;0) y (1;0)
  • 11. 2) Eje “y” cuando “x” es igual a cero. y = x2 -6x +5 Punto de corte: (0;5) y = 02 -6(0) +5 y = 0 - 0 +5 y = +5
  • 12. Puntos de corte: • “x” cuando “y” es igual a cero: (5;0) y (1;0) • “y” cuando “x” es igual a cero: (0;5) Ahora graficamos! Vértice (3 ;-4)