Answer:
Step-by-step explanation:
We only need 3 of those points and the standard form of the quadratic to find the function. The standard form is
[tex]y=ax^2+bx+c[/tex]
From each coordinate we will sub in the x and y values. I am going to use the following 3 points from the table. (0, 22), (1, 11) and (2, 2). Always start with a coordinate that has an x value of 0 if you can! Filling in the standard form with the first of those 3 coordinates:
[tex]22=a(0)^2+b(0)+c[/tex] simplifies down to the fact that c = 22. That's good to know; now we have a value for one of the variables to fill into the next substitution:
[tex]11=a(1)^2+b(1)+22[/tex]
which simplifies down to
a + b = -11
Now for the 3rd and last coordinate:
[tex]2=a(2)^2+b(2)+22[/tex]
which simplifies down to
4a + 2b = -20
Put the 2 bold equations together as a system and solve for one variable or the other. I chose to solve for b first, so in order to do that, I multiplied the first equation by a -4, giving me a new system:
-4a - 4b = 44
4a + 2b = -20
and
-2b = 24 so
b = -12. Now plug that back in and solve for a:
a - 12 = -11 so
a = 1.
Now we have the function:
[tex]y=x^2-12x+22[/tex]
