Question

In an experiment, a petri dish with a colony of bacteria is exposed to cold temperatures and then warmed again.

Answer 1

You have to solve 3 equations pretty much simultaneously here using the x and y values in the general form of the quadratic equation $y=a x^{2} +bx+c$
Start with the first values of x=0 and y=5.1 to solve for c:
$5.1=a(0) ^{2} +b(0)+c$ so c = 5.1
Next use the second x and y values along with the value of c in the next equation:
$3.03=a(1) ^{2} +b(1)+5.1$ gives you
-2.07 = a + b.  Solve this for a:
a = -2.07 - b
Finally use the third set of numbers with the c value AND the subbed value for a:
$1.17=a(3) ^{2} +b(3)+5.1$
1.17 = 9(-2.07 - b)+ 3b + 5.1 which simplifies to:
1.17 = -18.63 - 9b + 3b + 5.1 which further simplifies to:
14.7 = -6b and b = -2.45
Now you have c and b, let's find a using one of the simplified equations:
-2.07 = a + b
-2.07 = a - 2.45
a = .38
So here's your equation:
$y=.38 x^{2} -2.45x+5.1$