Find an equation in standard form of a parabola passing through the points below (1,-1),(5,31),(6,44).

Basic parabola:  y = ax2 + bx + c

We have 3 points we can plug in for (x, y) to create 3 simultaneous equations

(-2, 24):  24 = 4a - 2b + c   {equation 1}

(3, -1):   -1 = 9a + 3b + c    {equation 2}

(-1, 15): 15 = a - b + c        {equation 3}

Solve this system to find the values of a, b, c

Let's first eliminate variable c:

4a - 2b + c = 24   {equation 1}

 a -  b + c = 15    {equation 3}

 subtract

3a - b = 9

9a + 3b + c = -1    {equation 2}

 a -  b + c = 15    {equation 3}

 subtract

8a + 4b = -16

We now have two equations with 2 unknowns we can use to find a, b

3a - b = 9         {equation 4}

8a + 4b = -16   {equation 5}

Multiply equation 4 through by 4 and add equations

12a - 4b = 36

8a + 4b = -6

  add

20a = 30

a = 30/20

a = 3/2

8a + 4b = -6

8(3/2) + 4b = -6

12 + 4b = -6

4b = -6- 12

4b = -18

b = -18/4

b = -9/2

Plug these 2 values into one of the original equations and solve for c

15 = a - b + c        {equation 3}

15 = 3/2 + 9/2 + c

15 = 12/2 + c

15 = 6 + c

c = 15-6

c = 9

y = (3/2)x2 - (9/2)x+ 9

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Step-by-step explanation:

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9