The standard form of the quadratic equation is: [tex]y=ax^2+bx+c[/tex].
To find the standard form from our table, we are going to find the constant term, [tex]c[/tex], first. To do that we are going to look for the value of [tex]y[/tex] in or table in which [tex]x=0[/tex]. From our table we can infer that when [tex]x=0[/tex] [tex]y=-6[/tex], so [tex]c=-6[/tex].
So now we have: [tex]y=ax^2+bx-6[/tex]
Next, to find [tex]a[/tex] and [tex]b[/tex] we are going to use the points (1,2) and (3,0) from our table to create a system of equations:
- Using the point (1,2)
[tex][tex]y=ax^2+bx-6[/tex][/tex]
[tex]2=a(1)^2+b(1)-6[/tex]
[tex]2=a+b-6[/tex]
[tex]a+b=8[/tex] equation (1)
- Using the point (3,0):
[tex][tex]y=ax^2+bx-6[/tex][/tex]
[tex]0=a(3)^2+b(3)-6[/tex]
[tex]0=9a+3b-6[/tex]
[tex]9a+3b=6[/tex] equation (2)
Now we can solve our system of equations:
From equation (1)
[tex]a+b=8[/tex]
[tex]a=8-b[/tex] equation (3)
Replacing equation (3) in equation (2)
[tex]9a+3b=6[/tex]
[tex]9(8-b)+3b=6[/tex]
[tex]72-9b+3b=6[/tex]
[tex]72-6b=6[/tex]
[tex]-6b=-66[/tex]
[tex]b= \frac{-66}{-6} [/tex]
[tex]b=11[/tex] equation (4)
Replacing equation (4) in equation (3)
[tex]a=8-b[/tex]
[tex]a=8-11[/tex]
[tex]a=-3[/tex]
Putting all together:
[tex][tex]y=ax^2+bx-6[/tex][/tex]
[tex]y=-3x^2+11x-6[/tex]
We can conclude that the standard form of the quadratic equation from our table is: [tex]y=-3x^2+11x-6[/tex]
