Given a long algebraic equation, what are some strategies that you can use to make simplifying and evaluating the equation more
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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
using (0, 7):
gives you that c = 7. We will use that value now when we write the next 2 equations. Now the point (-2, 19):
and
so
12 = 4a - 2b
Now for the next point (-1, 12):
and
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