Question

1.Oblicz miejsca zerowe funkcji:

a) y= 2x²-14x+20

b) y= -⅓x + ⁴/₃ + -4/3

c) y= 6x² + x

d) y= -15x² + 3

e) y= -2(x+7)(x-12)

f) y= -5(x+1)² - 40

Answer 1

a. x₁ = 5 , x₂=2

b.x=⅓  ??

c. x₁=0, x₂=-⅙

d. x₁=-√5/5 , x₂=√5/5

e.x₁=12, x₂=-7

Δ<0 br. msc.zerowych

Answer 2

a) y= 2x²-14x+20

2x² -14x + 20 = 0

Δ= b² - 4ac = (-14)² - 4*2*20 = 196 – 160= 36

√Δ = √36= 6

x₁ = (-b-√Δ)/2a = (14-6)/4 = 8/4 = 2

x₂ = (-b+√Δ)/2a = (14+6)/4 = 20/4 = 5

 

c) y= 6x² + x

6x² + x = 0

x(6x +1) = 0

6x+ 1 = 0 lub x = 0 

6x= -1

x= -1/6

 

d) y= -15x² + 3

Δ= b² - 4ac = - 4*(-15)*3= 180

√Δ = √180 = √36*5 = 6√5

x₁ = -√Δ/2a = -6√5/-30 = √5/5

x₂ = √Δ/2a = 6√5/-30 =  -√5/5

 

e) y=  -2(x+7)(x-12)

-2(x +7)(x-12) = 0

-(x² -5x -84) = 0

-x² +5x +84 = 0

Δ= b² - 4ac = 5² - 4*(-1)*84= 25 + 336 = 361

√Δ = √361= 19

x₁ = (-b-√Δ)/2a = (-5-19)/-2 = -24/-2 = 12

x₂ = (-b+√Δ)/2a = (-5+19)/-2 = 14/-2 = -7

 

f) y= -5(x+1)² - 40

-5(x+1)² - 40 = 0

-5( x² + 2x +1) -40 = 0

-5 x² -10x -5 -40 = 0

-5x² -10x -45 = 0  /: (-5)

x² + 2x + 9 = 0

Δ= b² - 4ac = 2² - 4*1*9= 4- 36 = -32

nie ma miejsc  zerowych