Can anyone me i don’t know how to do this

Start with the vertex form of the equation of a quadratic. Fill in the numbers you know and solve for the one you don't know.
  y = a(x -h)² +k . . . . . . . . for some stretch factor "a" and vertex (h, k)

1. Vertex = (h, k) = (0, 0). Stretch factor can be found from the given point (x, y).
  8 = a(-2 -0)² +0
  8 = 4a . . . . . . . . simplify
  2 = a . . . . . . . . . divide by the coefficient of "a"
Your equation is y = 2x².

Note the above method for solving these problems. It repeats.

2. As above, substitute what's given and solve for what's not.
  3 = a(1 -2)² +0
  3 = a
Your equation is y = 3(x -2)²

3. Repeat
  -4 = a(-5 +3)² +0
  -4 = 4a
  -1 = a
Your equation is y = -(x +3)²

4. Are you seeing the pattern yet?
  0 = a(-1 -0)² +1
  0 = a +1
  -1 = a
Your equation is y = -x² +1

5 & 6. You know that if you want a zero (x-intercept) to be located at x=a, then a factor of the quadratic will be (x -a). Multiply the factors together to get your quadratic function.

5. y = (x -0)*(x -2) = x² -2x

6. y = (x +5)*(x -5) = x² -25 . . . . . . . . . . you should recognize this "special form", the factoring of the difference of squares

Num of plants = (original#ofplants)*(1.6)^t