Answer:
[tex]\displaystyle y = x^2 - 4x[/tex]
Step-by-step explanation:
To begin with, to define the stretch factour [tex]\displaystyle [A],[/tex]from the standard quadratic equation of [tex]\displaystyle y = Ax^2 + Bx + C,[/tex]find the rate of change between the vertex of [tex]\displaystyle [2, -4][/tex] and either [tex]\displaystyle [3, -3][/tex]or [tex]\displaystyle [1, -3].[/tex]It really does not matter which ordered pair you choose to pair with the vertex because one result will be negative, and the other result will be positive, so it is as though you are taking the absolute value of the rate of change, but not entirely. Anyhow, let us get to wourk:
[tex]\displaystyle \frac{-y_1 + y_2}{-x_1 + x_2} = m \\ \\ \frac{4 - 3}{-2 + 3} \hookrightarrow \frac{1}{1} \\ \\ 1 = m[/tex]
OR
[tex]\displaystyle \frac{-y_1 + y_2}{-x_1 + x_2} = m \\ \\ \frac{4 - 3}{-2 + 1} \hookrightarrow \frac{1}{-1} \\ \\ -1 = m[/tex]
So, both oppocite rate of changes tell you that the stretch factour of this quadratic function is [tex]\displaystyle 1,[/tex]ESPECIALLY sinse the graph opens upward.
Now, to determine the quadratic equation for this graph, you need to write this in vertex form first. Remember to base the equation off the vertex:
[tex]\displaystyle y = A[x - h]^2 + k \\ \\ y = [x + 3]^2 + 3[/tex]
So, −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, and sinse the vertex [tex]\displaystyle [(h, k)][/tex]is at [tex]\displaystyle [2, -4],[/tex]there MUST BE a plus sign inside those brackets. Anyway, let us continue:
[tex]\displaystyle y = [x - 2]^2 - 4 \hookrightarrow y = x^2 - 4x + 4 - 4 \\ \\ \boxed{y = x^2 - 4x}[/tex]
You now have your quadratic equation.
I am delighted to assist you at any time.
