Answer:
[tex]y=x^2+2x-8[/tex]
Step-by-step explanation:
When you graph those points on a piece of graph paper it appears that the points are in the form of a positive x^2 parabola, which has the standard form
[tex]y=ax^2+bx+c[/tex]
We just need to solve for a, b, and c. Easy. We have 3 points from the table. We will use all three of them to find the values of a, b, and c.
Use the points (0, -8), (2, 0), and (4, 16). You can use any points, but I chose the one with an x value of 0 for a good reason, and chose the other 2 because I don't like too many negatives!
Use the first point in those above to solve for c:
[tex]-8=a(0)^2+b(0)+c[/tex]
From this you solve for c: c = -8
Now use the next point along with the value of c to find another equation:
[tex]0=a(2)^2+b(2)-8[/tex] and
[tex]0=4a+2b-8[/tex] so
8 = 4a + 2b
That equation will be used again in a minute.
Use the last point to solve for yet another equation (stay with me...we are almost there!):
[tex]16=a(4)^2+b(4)-8[/tex] and
24 = 16a + 4b
Now we will use the method of elimination to solve for b:
8 = 4a + 2b
24 = 16a + 4b
Multiply the first equation by -4 to eliminate the a terms:
-32 = -16a - 8b
24 = 16a + 4b
leaves you with
-4b = -8 and b = 2. Now plug that back in to solve for a:
If 8 = 4a + 2b, then 8 = 4a + 2(2) and 8 = 4a + 4
4a = 4 and a = 1
Again, your equation is
[tex]y=x^2+2x-8[/tex]
