Answer:
Quadratic function that contains the points (0,-2), (1,0) and (3,10) is:
y=x^2+x-2
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