First, lets remenber some properties of de second degree polynomial.
Given a polynomial with the form:
[tex]y=ax^2+bx+c[/tex]
We can factor it as:
[tex]y=a(x-r_1)(x-r_2)[/tex]
where r_1 and r_2 are the roots of the polynomial.
Also, the polynomial "cuts" the Y axi in 'c', as follows:
[tex]y=a(0)^2+b(0)+c=c[/tex]
Or, in other words, in the point (0,c).
With those informations, we can solve our question.
From the graph, we can see that the roots are 1 and 5.
So, our polynomial would have the form (in the factored form):
[tex]y=a(x-1)(x-5)[/tex]
Expanding our expression, we have:
[tex]y=a(x-1)(x-5)\rightarrow y=ax^2-6ax+5a[/tex]
So, in our expression, c=5a. From the graph, we can see that the parabola cuts the Y in the point (0,5), so 'c' must be equal to 5. With that we can find the value of a, as follows:
[tex]c=5\rightarrow5a=5\rightarrow a=1[/tex]
So, our polynimial has the form:
[tex]y=x^2-6x+5\rightarrow y=(x^2-6x+\text{ 9) -4 }\rightarrow y=(x-3)^2-4[/tex]
