Find missing value for each quadratic function

I'll do the first one to get you started. The answer is -3

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1

0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

y = -4 + 1

y = -3

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

50

step-by-step explanation: