Create the quadratic function that contains the points (0, -2), (1, 0) and (3, 10). show all of your work for full credit.

Quadratic function that contains the points (0,-2), (1,0) and (3,10) is:

y=x^2+x-2

Step-by-step explanation:

y=ax^2+bx+c

Point 1: (0,-2)=(x,y)→x=0, y=-2

Replacing in the equation:

-2=a(0)^2+b(0)+c→-2=a(0)+0+c→-2=0+0+c→-2=c→c=-2

y=ax^2+bx+c

c=-2→y=ax^2+bx+(-2)→y=ax^2+bx-2

Point 2: (1,0)=(x,y)→x=1, y=0

Replacing in the equation:

0=a(1)^2+b(1)-2→0=a(1)+b-2→0=a+b-2

Adding 2 both sides of the equation:

0+2=a+b-2→2=a+b→a+b=2 (Equation 1)

Point 3: (3,10)=(x,y)→x=3, y=10

Replacing in the equation:

10=a(3)^2+b(3)-2→10=a(9)+3b-2→10=9a+3b-2

Adding 2 both sides of the equation:

10+2=9a+3b-2+2→12=9a+3b→9a+3b=12 (Equation 2)

We have the following system of 2 equations (Equation 1 and Equation 2) and 2 unkowns (a and b):

Eq. 1: a+b=2

Eq. 2: 9a+3b=12

Solving the system of equations using the Method of Substitution:

Isloating b from Equation 1: Subtracting a from both sides of the equation:

Eq. 1: a+b-a=2-a→b=2-a

Replacing b by 2-a in Equation 2:

Eq. 2: 9a+3b=12→9a+3(2-a)=12

Eliminating the parentheses applying the distribution property:

9a+3(2)+3(-a)=12

Multiplying:

9a+6-3a=12

Adding like terms:

6a+6=12

Solving for a: Subtracting 6 from both sides of the equation:

6a+6-6=12-6

Subtracting:

6a=6

Dividing both sides of the equation by 6:

6a/6=6/6

Dividing:

a=1

Replacing a by 1 in the equation b=2-a

b=2-1→b=1

Replacing a and b in the quadratic equation:

y=ax^2+bx-2

y=1 x^2+1 x-2

y=x^2+x-2

First change each value into improper fractions, which is 31/2 and 29/8. To divide fractions, keep the first one the same, change to multiplication, and flip the 29/8. So, the equation is now 31/2 * 8/29. Cross cancel the 2 & 8, making the equation 31/1 * 4/29. Multiply 31 by 4 to get the numerator, 124. Then divide 124 by 29 to get the final answer, ~4.27... So, she will need 5 buckets.