Evaluate each function at each of the four points.

Functions:
f(x) = x2 – 5x – 6
f(x) = 23 – 22 – 12
f(x) = 5.24

Points:
f(0)
f(2)
f(-1)
f(6)

1) f(0)= -6 , f(2)=  -12, f(-1)=  0, f(6)= 0

2) f(0)= -12 , f(2)= -8 , f(-1)= -14 , f(6)= 168

3) f(0)= 5 , f(2)=  20, f(-1)= 2.5 , f(6)= 320

Step-by-step explanation:

1) The given function is f(x) = x² – 5x – 6.

To find the values of  f(0) , f(2),  f(-1),  f(6) :

Substitute x=0 in f(x) = x² – 5x – 6

f(0) = -6

Therefore, f(0) = -6

Substitute x=2 in f(x) = x² – 5x – 6

f(2)  = (2)² -5(2)-6

⇒ 4-10-6

⇒ 4-16

⇒ -12

Therefore, f(2) = -12

Substitute x= -1 in f(x) = x² – 5x – 6

f(-1)  = (-1)² - 5(-1) -6

⇒  1+5-6

⇒ 6-6

⇒ 0

Therefore, f(-1)  = 0

Substitute x= 6 in f(x) = x² – 5x – 6

f(6) = (6)² - 5(6) -6

⇒ 36 -30 -6

⇒ 36-36

⇒ 0

Therefore, f(6) = 0

2) The given function is f(x) = x³– x²–12.

To find the values of  f(0) , f(2),  f(-1),  f(6) :

Substitute x=0 in f(x) = x³– x²–12

f(0) = -`12

Therefore, f(0) = -12

Substitute x= 2 in f(x) = x³– x²–12

f(2)  = (2)³ - (2)² -12

⇒ 8-4-12

⇒ 8-16

⇒ -8

Therefore, f(2)  = -8

Substitute x= -1 in f(x) = x³– x²–12

f(-1)  = (-1)³ - (-1)² -12

⇒  -1-1-12

⇒ -14

Therefore, f(-1)  = -14

Substitute x= 6 in f(x) =  x³– x²–12

f(6) = (6)³ - (6)² -12

⇒ 216 -36 -12

⇒ 216-48

⇒ 168

Therefore, f(6) = 168

3) The given function is f(x) =

To find the values of  f(0) , f(2),  f(-1),  f(6) :

Substitute x=0 in f(x) =

f(0) = 5 ×

Any number with power zero is 1.

f(0) = 5

Therefore, f(0) = 5

Substitute x=2 in f(x) =

f(2) = 5 × 2²

⇒ 5×4

⇒ 20

Therefore, f(2) = 20

Substitute x= -1 in f(x) =

f(-1)  =

⇒ 5 × 1/2

⇒ 5 × 0.5

⇒ 2.5

Therefore, f(-1)  = 2.5

Substitute x= 6 in f(x) =  

f(6) =

⇒ 5 × 64

⇒ 320

Therefore, f(6) = 320

answer: 0.64

step-by-step explanation:

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.