The given graph has 1 and 3 as zeroes of the function, the graph is also parabolic downwards, so we know that the leading coefficient is negative. It's vertex is at (2,1)
[tex]\begin{gathered} f(x)=-(x-1)(x-3)\text{ based on the roots} \\ f(x)=-(x-2)^2+1\text{ based on the vertex} \\ f(x)=-x^2+4x-3\text{ if we simplify }-(-x-1)(x-3) \\ \end{gathered}[/tex][tex]\begin{gathered} \text{The standard form of a quadratic equation is }f(x)=a(x-h)^2+k \\ \text{where }(h,k)\text{ is the vertex} \end{gathered}[/tex]
By inspection of the graph, the vertex is found at (2,1), since it is facing downwards, it also gives us an idea that a is negative.
[tex]\begin{gathered} \text{The function} \\ f(x)=-(x-2)^2+1 \\ \text{is already on the the vertex form, and it tells us that }a\text{ is negative because it is preceeded} \\ \text{by a negative sign, and that the vertex is at }(2,1)\text{ which is found on the function.} \end{gathered}[/tex][tex]\begin{gathered} \text{The zeroes, or the x-intercept, are values of }x\text{ for which the function crosses the x-axis} \\ \text{Inspecting the graph, it crosses at }x=1,\text{ and }x=3 \\ \text{Equate these to zero and we get} \\ x=1 \\ x-1=0 \\ \\ x=3 \\ x-3=0 \\ \\ \text{which means that }(x-1)\text{ and }(x-3)\text{ are factors, and since it is facing downward it} \\ \text{will be preceeded by a negative sign hence} \\ f(x)=-(x-1)(x-3) \end{gathered}[/tex]
