Answer:
[tex]\begin{gathered} 0.44x^{2}-2.67x+3 \\ or \\ \frac{1}{2.25}(x-3)^2-1 \end{gathered}[/tex]
Explanation:
From the graph, we will observe that the vertex is at the point:
[tex]\begin{gathered} (h,k)=(3,-1) \\ \text{We will obtain the quadratic equation from the vertex form using the formula:} \\ f(x)=a(x-h)^2+k \\ \text{Inputting the values of ''h'' \& ''k'', we have:} \\ f(x)=a(x-3)^2-1 \\ \text{We have the x-intercept at point:} \\ (x,y)=(1.5,0) \\ f(1.5)=0 \\ \Rightarrow0=a(1.5-3)^2-1 \\ 0=a(-1.5)^2-1 \\ 0=a(2.25)-1 \\ 0=2.25a-1 \\ \text{Add ''1'' to both sides, we have:} \\ 1=2.25a \\ 2.25a=1 \\ \text{Divide both sides by ''2.25'', we have:} \\ a=\frac{1}{2.25}=0.44 \\ \text{We will substitute the value of ''a'' into the vertex equation, we have:} \\ f(x)=\frac{1}{2.25}(x-3)^2-1 \\ \text{Expand the bracket, we have:} \\ f(x)=\frac{1}{2.25}(x-3)(x-3)-1 \\ f(x)=\frac{1}{2.25}[x(x-3)-3(x-3)]-1 \\ f(x)=\frac{1}{2.25}[x^2-3x-3x+9]-1 \\ f(x)=\frac{1}{2.25}[x^2-6x+9]-1 \end{gathered}[/tex]
We will obtain the quadratic function as shown below:
[tex]\begin{gathered} \frac{1}{2.25}[x^2-6x+9]-1 \\ \text{Expand the bracket, we have:} \\ 0.44x^2-2.67x+4-1 \\ 0.44x^2-2.67x+3 \\ or \\ \frac{1}{2.25}(x-3)^{2}-1 \\ \end{gathered}[/tex]
