Match each function formula with the corresponding transformation of the parent function y = –x 2 – 1. 1. y = –x2 – 1 Translated right by 1 unit 2. y = –(x – 1)2 – 1 Reflected across the x-axis 3. y = x2 + 1 Translated down by 1 unit 4. y = –x2 Reflected across the y-axis 5. y = –(x + 1)2 – 1 Translated left by 1 unit 6. y = –x2 – 2 Translated up by 1 unit

1. f₁(x)=-x²+2x-2

2. f₂(x)=-x²+6x-10

3. f₃(x)=-x²-2  

4. f₄(x)=-x²+10x-26

5. f₅(x)=-x²-2x-2

6. f₆(x)=-x²

Step-by-step explanation:

1. we have that f(x) is translated right by 1 unit , also

according to the graph f(x) has a vertex P(0,-1) then P₁ ( 1,-1) to f₁(x) therefore y-y₁ = -(x-x₁)² ⇒ y-y₁ = -(x-x₁)² ⇒ y+1 = -(x-1)² ⇒ y=-(x²-2x+1)-1 =-x²+2x-2

2. f(x) is reflected across the x-axis 3, same as in 1., P₂(3,-1) to f₂(x)

y+1 = -(x-3)² ⇒ y=-(x²-6x+9)-1 =-x²+6x-10

3. f(x) is translated down by 1 unit, P₃(0,-2) to f₃(x)

y+2 = -(x)² ⇒ y=-x²-2  

4. f(x) is reflected across the y-axis 5, P₄(5,-1) to f₄(x)

y+1 = -(x-5)² ⇒ y=-(x²-10x+25)-1 =-x²+10x-26

5. f(x) is translated left by 1, P₅(-1,-1) to f₅(x)

y+1 = -(x+1)² ⇒ y=-(x²+2x+1)-1 =-x²-2x-2

6. f(x) is translated up by 1 unit, P₆(0,0) to f₆(x)

y+0 = -(x+0)² ⇒ y=-x²

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: