How would you find the equation that goes hits all three of these points?

Question

The points are: (1,4), (4,22), and (10,112)
The equation is: ?

Thanks!

Answer 1

Well, you can get a quadratic equation to go through any three points.

The quadratic equation is always y=ax^2+bx+c. From this you can find out the slope of a secant line like so:

[a(x+h)^2+b(x+h)+c - (ax^2 + by + c)]/(x+h - x)
[ax^2+2axh+ah^2 + bx+bh + c - ax^2 - bx - c)/h
(2axh + ah^2 + bh)/h
2ax + ah + b
a(2x+h) + b

So lets take what we know, when x=1 and h=3, the slope is 6 and when x=4 and h=6 then the slopes is 15

5a+b = 6
14a+b = 15

Then you can solve for a and b

6-5a=15-14a

9a=9
a=1
b=6-5a
b=1

So now we have y = x^2+x+c and when x = 1, y = 4.

4 = 2 + c
c = 2

So the equation is:

y = x^2+x+2

Answer 2

x^2 + x + 2.

Honestly, I'm not sure how to work that out. The fact that it's increasing so fast gave me the clue that it was x^2. I looked at the 10,112 and saw that at least another 10 had to get added, so I tacked on an X. Then just put the 2 on there and got lucky that it worked with the other two points as well.