Answer:
Step-by-step explanation:
You have 3 unknowns: a, b, and c. It's our job to find them algebraically. I'm going to start with the point where x = 0 and y = 7. You'll see why in a minute. Filling in the standard form of a quadratic
[tex]y=ax^2+bx+c[/tex] using (0, 7):
[tex]7=a(0)^2+b(0)+c[/tex] gives you that c = 7. We will use that value now when we write the next 2 equations. Now the point (-2, 19):
[tex]19=a(-2)^2+b(-2)+7[/tex] and
[tex]19=4a-2b+7[/tex] so
12 = 4a - 2b
Now for the next point (-1, 12):
[tex]12=a(-1)^2+b(-1)+7[/tex] and
[tex]12=a-b+7[/tex] so
5 = a - b
Now we have a system of equations (the 2 bold font equations) that we will solve by elimination:
12 = 4a - 2b
5 = a - b
Multiply the bottom equation by -4 to get a new system:
12 = 4a - 2b
-20 = -4a + 4b
Add those together to get rid of the a terms and end up with
-8 = 2b so
b = -4
Now we can sub in -4 for b to solve for a. I'm using the second bold type equation to do this:
5 = a - (-4) and
5 = a + 4 so
a = 1 and the equation for the quadratic function is
[tex]y=x^2-4x+7[/tex]
