x 1 2 3 4
h(x) 7 20 43 76
x = 1 h(x) = 7
x = 2 h(x) = 20
x = 3 h(x) = 43
x = 4 h(x) = 76
Using the table or in this case information for the polynomial function of the degree 2
Find the formula for h(x)
h(x) = ?
There's no guarantee that a single parabola will pass through all four points. So, let's pick the fitrst three and see what happens.
If y = ax^2+bx+c, then
a+b+c = 7
4a+2b+c = 20
9a+3b+c = 43
3a+b = 13
5a+b = 23
2a = 10
a = 5
b = -2
c = 4
y = 5x^2-2x+4
Luckily, (4,76) also fits the equation.